# Enumerate the Phat-fingered-lights-out numbers

Even though the concept of phat-fingered-lights-out number should be pretty self-explanatory here is a definition:

Given a nonnegative integer in binary representation a phat-fingered double-bit-flip or phatflip for short replaces a pair of adjacent, equal bits with their complement.

For example given the number 0b111010010000101

0b111010010000101 -> 0b100010010000101

0b111010010000101 -> 0b111011110000101

0b111010010000101 -> 0b1010010000101

are phatflips whereas

0b111010010000101 -> 0b111010001000101 (not equal)

0b111010010000101 -> 0b111010010111101 (not a pair)

are not.

A phat-fingered-lights-out number (or "pflon") is a nonnegative integer that can be completely erased by a finite sequence of phatflips.

The first 1000 pflons are:

0, 3, 6, 9, 12, 15, 18, 24, 27, 30, 33, 36, 39, 45, 48, 51, 54, 57, 60, 63, 66, 72, 75, 78, 90, 96, 99, 102, 105, 108, 111, 114, 120, 123, 126, 129, 132, 135, 141, 144, 147, 150, 153, 156, 159, 165, 177, 180, 183, 189, 192, 195, 198, 201, 204, 207, 210, 216, 219, 222, 225, 228, 231, 237, 240, 243, 246, 249, 252, 255, 258, 264, 267, 270, 282, 288, 291, 294, 297, 300, 303, 306, 312, 315, 318, 330, 354, 360, 363, 366, 378, 384, 387, 390, 393, 396, 399, 402, 408, 411, 414, 417, 420, 423, 429, 432, 435, 438, 441, 444, 447, 450, 456, 459, 462, 474, 480, 483, 486, 489, 492, 495, 498, 504, 507, 510, 513, 516, 519, 525, 528, 531, 534, 537, 540, 543, 549, 561, 564, 567, 573, 576, 579, 582, 585, 588, 591, 594, 600, 603, 606, 609, 612, 615, 621, 624, 627, 630, 633, 636, 639, 645, 657, 660, 663, 669, 693, 705, 708, 711, 717, 720, 723, 726, 729, 732, 735, 741, 753, 756, 759, 765, 768, 771, 774, 777, 780, 783, 786, 792, 795, 798, 801, 804, 807, 813, 816, 819, 822, 825, 828, 831, 834, 840, 843, 846, 858, 864, 867, 870, 873, 876, 879, 882, 888, 891, 894, 897, 900, 903, 909, 912, 915, 918, 921, 924, 927, 933, 945, 948, 951, 957, 960, 963, 966, 969, 972, 975, 978, 984, 987, 990, 993, 996, 999, 1005, 1008, 1011, 1014, 1017, 1020, 1023, 1026, 1032, 1035, 1038, 1050, 1056, 1059, 1062, 1065, 1068, 1071, 1074, 1080, 1083, 1086, 1098, 1122, 1128, 1131, 1134, 1146, 1152, 1155, 1158, 1161, 1164, 1167, 1170, 1176, 1179, 1182, 1185, 1188, 1191, 1197, 1200, 1203, 1206, 1209, 1212, 1215, 1218, 1224, 1227, 1230, 1242, 1248, 1251, 1254, 1257, 1260, 1263, 1266, 1272, 1275, 1278, 1290, 1314, 1320, 1323, 1326, 1338, 1386, 1410, 1416, 1419, 1422, 1434, 1440, 1443, 1446, 1449, 1452, 1455, 1458, 1464, 1467, 1470, 1482, 1506, 1512, 1515, 1518, 1530, 1536, 1539, 1542, 1545, 1548, 1551, 1554, 1560, 1563, 1566, 1569, 1572, 1575, 1581, 1584, 1587, 1590, 1593, 1596, 1599, 1602, 1608, 1611, 1614, 1626, 1632, 1635, 1638, 1641, 1644, 1647, 1650, 1656, 1659, 1662, 1665, 1668, 1671, 1677, 1680, 1683, 1686, 1689, 1692, 1695, 1701, 1713, 1716, 1719, 1725, 1728, 1731, 1734, 1737, 1740, 1743, 1746, 1752, 1755, 1758, 1761, 1764, 1767, 1773, 1776, 1779, 1782, 1785, 1788, 1791, 1794, 1800, 1803, 1806, 1818, 1824, 1827, 1830, 1833, 1836, 1839, 1842, 1848, 1851, 1854, 1866, 1890, 1896, 1899, 1902, 1914, 1920, 1923, 1926, 1929, 1932, 1935, 1938, 1944, 1947, 1950, 1953, 1956, 1959, 1965, 1968, 1971, 1974, 1977, 1980, 1983, 1986, 1992, 1995, 1998, 2010, 2016, 2019, 2022, 2025, 2028, 2031, 2034, 2040, 2043, 2046, 2049, 2052, 2055, 2061, 2064, 2067, 2070, 2073, 2076, 2079, 2085, 2097, 2100, 2103, 2109, 2112, 2115, 2118, 2121, 2124, 2127, 2130, 2136, 2139, 2142, 2145, 2148, 2151, 2157, 2160, 2163, 2166, 2169, 2172, 2175, 2181, 2193, 2196, 2199, 2205, 2229, 2241, 2244, 2247, 2253, 2256, 2259, 2262, 2265, 2268, 2271, 2277, 2289, 2292, 2295, 2301, 2304, 2307, 2310, 2313, 2316, 2319, 2322, 2328, 2331, 2334, 2337, 2340, 2343, 2349, 2352, 2355, 2358, 2361, 2364, 2367, 2370, 2376, 2379, 2382, 2394, 2400, 2403, 2406, 2409, 2412, 2415, 2418, 2424, 2427, 2430, 2433, 2436, 2439, 2445, 2448, 2451, 2454, 2457, 2460, 2463, 2469, 2481, 2484, 2487, 2493, 2496, 2499, 2502, 2505, 2508, 2511, 2514, 2520, 2523, 2526, 2529, 2532, 2535, 2541, 2544, 2547, 2550, 2553, 2556, 2559, 2565, 2577, 2580, 2583, 2589, 2613, 2625, 2628, 2631, 2637, 2640, 2643, 2646, 2649, 2652, 2655, 2661, 2673, 2676, 2679, 2685, 2709, 2757, 2769, 2772, 2775, 2781, 2805, 2817, 2820, 2823, 2829, 2832, 2835, 2838, 2841, 2844, 2847, 2853, 2865, 2868, 2871, 2877, 2880, 2883, 2886, 2889, 2892, 2895, 2898, 2904, 2907, 2910, 2913, 2916, 2919, 2925, 2928, 2931, 2934, 2937, 2940, 2943, 2949, 2961, 2964, 2967, 2973, 2997, 3009, 3012, 3015, 3021, 3024, 3027, 3030, 3033, 3036, 3039, 3045, 3057, 3060, 3063, 3069, 3072, 3075, 3078, 3081, 3084, 3087, 3090, 3096, 3099, 3102, 3105, 3108, 3111, 3117, 3120, 3123, 3126, 3129, 3132, 3135, 3138, 3144, 3147, 3150, 3162, 3168, 3171, 3174, 3177, 3180, 3183, 3186, 3192, 3195, 3198, 3201, 3204, 3207, 3213, 3216, 3219, 3222, 3225, 3228, 3231, 3237, 3249, 3252, 3255, 3261, 3264, 3267, 3270, 3273, 3276, 3279, 3282, 3288, 3291, 3294, 3297, 3300, 3303, 3309, 3312, 3315, 3318, 3321, 3324, 3327, 3330, 3336, 3339, 3342, 3354, 3360, 3363, 3366, 3369, 3372, 3375, 3378, 3384, 3387, 3390, 3402, 3426, 3432, 3435, 3438, 3450, 3456, 3459, 3462, 3465, 3468, 3471, 3474, 3480, 3483, 3486, 3489, 3492, 3495, 3501, 3504, 3507, 3510, 3513, 3516, 3519, 3522, 3528, 3531, 3534, 3546, 3552, 3555, 3558, 3561, 3564, 3567, 3570, 3576, 3579, 3582, 3585, 3588, 3591, 3597, 3600, 3603, 3606, 3609, 3612, 3615, 3621, 3633, 3636, 3639, 3645, 3648, 3651, 3654, 3657, 3660, 3663, 3666, 3672, 3675, 3678, 3681, 3684, 3687, 3693, 3696, 3699, 3702, 3705, 3708, 3711, 3717, 3729, 3732, 3735, 3741, 3765, 3777, 3780, 3783, 3789, 3792, 3795, 3798, 3801, 3804, 3807, 3813, 3825, 3828, 3831, 3837, 3840, 3843, 3846, 3849, 3852, 3855, 3858, 3864, 3867, 3870, 3873, 3876, 3879, 3885, 3888, 3891, 3894, 3897, 3900, 3903, 3906, 3912, 3915, 3918, 3930, 3936, 3939, 3942, 3945, 3948, 3951, 3954, 3960, 3963, 3966, 3969, 3972, 3975, 3981, 3984, 3987, 3990, 3993, 3996, 3999, 4005, 4017, 4020, 4023, 4029, 4032, 4035, 4038, 4041, 4044, 4047, 4050, 4056, 4059, 4062, 4065, 4068, 4071, 4077, 4080, 4083, 4086, 4089, 4092, 4095, 4098, 4104, 4107, 4110, 4122, 4128, 4131, 4134, 4137, 4140, 4143, 4146, 4152, 4155, 4158, 4170, 4194, 4200, 4203, 4206, 4218, 4224, 4227, 4230, 4233, 4236, 4239, 4242, 4248, 4251, 4254, 4257, 4260, 4263, 4269, 4272, 4275, 4278, 4281, 4284, 4287, 4290, 4296, 4299, 4302, 4314, 4320, 4323, 4326, 4329, 4332, 4335, 4338, 4344, 4347, 4350, 4362, 4386, 4392, 4395, 4398, 4410, 4458, 4482, 4488, 4491, 4494, 4506, 4512, 4515, 4518, 4521, 4524, 4527, 4530, 4536


This is code-golf and asequence challenge; standard rules apply:

• shortest code in bytes (full program or function) wins; each language or major language version is their own competition
• you can either take a number n and return the n-th pflon or the first n pflons; counting can be 0- or 1-based
• or you can take no input and output the sequence til kingdom come

Source: This is mostly a rip-off of this post over at puzzling SO.

• Is this A039004? Feb 17 at 16:57
• @Command Master, yes it is. I will include a proof in my answer.
– Tbw
Feb 17 at 17:06
• Why is is it "phat"? Feb 18 at 18:54
• @WheatWizard mostly me being facetious: picture someone with an unprecise touch operating a tiny array of switches. The spelling is to make it look cool as opposed to carrying stigma. Feb 18 at 20:44

# Uiua 0.9.0, 14 bytes SBCS (updated!)

↙:▽=0/-⍉⋯.⇡×..


Try on Uiua Pad! Takes $$\n\$$ as input and outputs the first $$\n\$$ terms. Does essentially the same thing as my answer below and @Mukundan314's APL answer.

## 19 bytes SBCS

⍥(+1(&p.|)≠0/-⋯.)∞0


Try it! Takes no input and outputs the sequence (change the ∞ to some small number like 10  if you want to see the beginning)

## Old Explanation

0 # push 0 to stack
⍥(       # repeat
⋯.     # to binary
/-     # alternating difference
≠0     # not zero?
(&p.|) # else print number
)∞       # infinitely many times


## Theorem

$$\n\$$ is a pflon if and only if in the binary representation of $$\n\$$, the number of 1s in even positions of is equal to the number of 1s in odd positions.

### Proof

($$\\implies\$$) Every phatflip clearly maintains whether the odd-even condition holds, as it either adds a 1 to both counts or removes a 1 from both counts. The condition holds for 0. Since any pflon can be obtained from phatflips starting at 0 (just reversing the sequence to get lights out), the condition holds for every pflon.

($$\\impliedby\$$) We prove by strong induction on $$\n\$$. The base case 0 is trivial.

Let $$\n>0\$$. Suppose all integers $$\k\$$ with $$\0 \leq k < n\$$ for which the binary condition holds are pflons. Then suppose the binary condition holds for $$\n\$$.

$$\n\$$ must have at least a 1 in both even and odd position, so it must have a 1 in even position and a 1 in odd position separated by zero or more 0s. Since their positions have opposite parity, there must be an even number of 0s between them. So, we can flip all those 0s into 1s, then flip all those and the two 1s back into zeros. All these flips maintain the binary condition, as shown above, and the result $$\k\$$ is less than $$\n\$$ as we just switched two 1s to 0s. Therefore, $$\k\$$ is a pflon, and is accessible from $$\n\$$ by phatflips, so $$\n\$$ is a pflon.

• Well explained proof, nice insight. Feb 17 at 18:25
• Much more thorough than my explanation! Feb 17 at 18:56
• ⍥(...)∞ means "apply a function until the input stops changing" now. If you want an infinite loop, you will need to say ⍢(...)1. Feb 17 at 19:28
• @chunes thanks, I didn't know that! In this case it doesn't really matter, and repeat saves a byte because you don't need a space between inf and 0.
– Tbw
Feb 18 at 0:30

# Jelly, 6 bytes

Bḅ-¬µ#


A full program that accepts a positive integer, n, on stdin and prints the first n phat-fingered-lights-out numbers.

Try it online!

### How?

Every phat-flip applies to an adjacent pair of bits, so if the number of odd-positioned ons is equal to the number of even-position ons then after flipping as many on-pairs as possible we can bring any remaining separated ons together by flipping pairs of offs and then flip remaining pairs of ons out of existence. A binary number with this property, is zero when converted from base $$\-1\$$ since, counting from the least significant bit as position $$\1\$$, odd-position ons contribute $$\1\$$ while even-position ons contribute $$\-1\$$ (and all offs contribute $$\0\$$).

Bḅ-¬µ# - Main Link: no arguments
µ# - count up from k=0, and yield the first {stdin} integers for which:
B      -   convert to binary
ḅ-    -   convert from base -1
¬   -   logical NOT
- implicit print


# JavaScript (V8), 45 bytes

A full program that prints the sequence indefinitely.

This relies on arithmetic underflow to stop the recursion, making it slower than it should (+1 byte for a faster version).

for(n=-1;f=n=>n&&n%2-f(n/2);)f(++n)||print(n)


Try it online!

# Charcoal, 2319 16 bytes

ＮθＩ…Φ×θθ¬↨↨ι²±¹θ


Try it online! Link is to verbose version of code. Outputs the first n terms. Explanation: Even more wildly overestimates an upper bound for A039004(n), then uses @Tbw's theorem to find n terms.

Ｎθ                  Input n as an integer
θ             Input n
×              Multiplied by
θ            Input n
Φ               Filter over implicit range where
ι        Current value
↨         Converted to base
²       Literal integer 2
↨          Interpreted as base
¹     Literal integer 1
±      Negated
¬           Is zero
…                Truncated to length
θ    Input n
Ｉ                 Cast to string
Implicitly print


Edit: Removed ⁺³Ｌ saving 3 bytes to even more wildly overestimate A039004(n) thanks to @Tbw.

Outputting all of the pflons less than n would of course be shorter:

ＩΦＮ¬↨↨ι²±¹


Try it online! Link is to verbose version of code. Explanation:

  Ｎ         Input n as a number
Φ          Filter over implicit range where
ι     Current value
↨      Converted to base
²    Literal integer 2
↨       Interpreted as base
¹  Literal integer 1
±   Negated
¬        Is zero
Ｉ           Cast to string
Implicitly print

• You can save at least 3 bytes since $n^2$ is a good enough upper bound
– Tbw
Feb 18 at 2:17
• @Tbw I guess I was overthinking it... the annoying thing is that it's much easier to estimate n given A039004(n) (not that you need it since it's also easier to calculate directly anyway).
– Neil
Feb 18 at 4:38

# APL(Dyalog Unicode), 18 bytes SBCS

Requires Index Origin 0 (⎕­⁡​‎‎⁪⁡⁪⁠⁪⁤⁤⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁡⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁡⁢⁪‏‏​⁡⁠⁡‌⁢​‎‎⁪⁡⁪⁠⁪⁤⁢⁪‏⁠‎⁪⁡⁪⁠⁪⁤⁣⁪‏‏​⁡⁠⁡‌⁣​‎‎⁪⁡⁪⁠⁪⁤⁡⁪‏‏​⁡⁠⁡‌⁤​‎‎⁪⁡⁪⁠⁪⁢⁤⁪‏⁠‎⁪⁡⁪⁠⁪⁣⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁣⁢⁪‏⁠‎⁪⁡⁪⁠⁪⁣⁣⁪‏⁠‎⁪⁡⁪⁠⁪⁣⁤⁪‏‏​⁡⁠⁡‌⁢⁡​‎‎⁪⁡⁪⁠⁪⁢⁢⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁣⁪‏⁠⁪⁪‏​⁡⁠⁡‌⁢⁢​‎⁪⁪⁠‎⁪⁡⁪⁠⁪⁤⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁡⁪‏‏​⁡⁠⁡‌⁢⁣​‎‎⁪⁡⁪⁠⁪⁣⁪‏‏​⁡⁠⁡‌⁢⁤​‎‎⁪⁡⁪⁠⁪⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁪‏‏​⁡⁠⁡‌­IO←0)

A↑⍸0=-⌿2⊥⍣¯1⍳×⍨A←⎕


Try it on APLgolf!

A full program which outputs the first n phat-fingered-lights-out numbers.

A↑⍸0=-⌿2⊥⍣¯1⍳×⍨A←⎕­⁡​‎‎⁪⁡⁪⁠⁪⁤⁤⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁡⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁡⁢⁪‏‏​⁡⁠⁡‌⁢​‎‎⁪⁡⁪⁠⁪⁤⁢⁪‏⁠‎⁪⁡⁪⁠⁪⁤⁣⁪‏‏​⁡⁠⁡‌⁣​‎‎⁪⁡⁪⁠⁪⁤⁡⁪‏‏​⁡⁠⁡‌⁤​‎‎⁪⁡⁪⁠⁪⁢⁤⁪‏⁠‎⁪⁡⁪⁠⁪⁣⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁣⁢⁪‏⁠‎⁪⁡⁪⁠⁪⁣⁣⁪‏⁠‎⁪⁡⁪⁠⁪⁣⁤⁪‏‏​⁡⁠⁡‌⁢⁡​‎‎⁪⁡⁪⁠⁪⁢⁢⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁣⁪‏⁠⁪⁪‏​⁡⁠⁡‌⁢⁢​‎⁪⁪⁠‎⁪⁡⁪⁠⁪⁤⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁡⁪‏‏​⁡⁠⁡‌⁢⁣​‎‎⁪⁡⁪⁠⁪⁣⁪‏‏​⁡⁠⁡‌⁢⁤​‎‎⁪⁡⁪⁠⁪⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁪‏‏​⁡⁠⁡‌­
A←⎕  # ‎⁡A = input
×⍨     # ‎⁢square
⍳       # range [0, 1, ..., arg - 1]
2⊥⍣¯1        # ‎⁤to base 2
-⌿             # ‎⁢⁡alternating difference
0=               # ‎⁢⁢equals zero
⍸                 # ‎⁢⁣indices of true
A↑                  # ‎⁢⁤take first A elements


# x86 32-bit machine code, 23 bytes

83 C8 FF 40 BA AA AA AA AA 31 C2 F3 0F B8 D2 83 FA 10 75 EF E2 ED C3


Try it online!

Following the fastcall calling convention, this takes a number n in ECX and returns the 1-indexed nth pflon in EAX.

In assembly:

f:  or eax, -1      # Set EAX to -1 (all 1 bits).
r:  inc eax         # Add 1 to EAX. EAX will be the number being tested.
mov edx, 0xAAAAAAAA # Set EDX to this value, with 1 bits in odd positions.
xor edx, eax    # Exclusive-OR that value with EAX (inverts odd positions).
popcnt edx, edx # Count the number of 1 bits in that value.
cmp edx, 16     # Check whether it's 16.
jne r           # If not, jump back (to test the next number).
loop r          # Decrease ECX by 1 and jump back if it's nonzero.
ret             # Return. (EAX is the return value.)


# J, 23 21 bytes

{.(#~0=-/"1@#:)@i.@*:


Try it online!

Uses Tbw's observation and is basically a port of Mukundan314's APL answer

# APL+WIN, 29 37 bytes

+8 bytes as it appears my original answer of outputting the series up to a given value was outside the rules. Prompts for number of values to output. Index origin = 0

n↑(0=-⌿((⌈2⍟1⌈n*2)⍴2)⊤m)/m←⍳1+(n←⎕)*2


Try it online! Thanks to Dyalog Classic

Prompts for an integer and outputs the sequence up to that value. Index origin = 0

(0=-⌿((⌈2⍟1⌈n)⍴2)⊤m)/m←⍳1+n←⎕

• I don't think it's allowed to output values till n for sequence problems.Allowed formats are: infinite sequence, nth term, first n terms Feb 18 at 5:41
• @Mukundan314. Thanks. Edited to comply with rules as stated. Feb 18 at 8:38

# Haskell + hgl, 20 bytes

fl(qb sm<%uak<bs2)nn


Attempt This Online!

## Explanation

We convert the list to base 2, split it into digits in an even position and digits in an odd position, then check that these two lists have the same sum.

Since 0 and 3 in base 4 are 00 and 11 in base 2, each of these contributes the same amount to both lists and have no effect on the outcome. 1 and 2 in base 4 are 01 and 10 in base 2, so each contributes 1 to one of sums of the two halves. Thus checking if the sums are the same as checking if there are the same number of 1s as 2s in the base 4 representation.

fl(eq0<rF(-)0<bs2)nn


Attempt This Online!

## Explanation

This uses the more complicated alternate sum approach found in this answer for the same byte count.

fl((ce1*=*ce2)<bs4)nn


Attempt This Online!

## Explanation

This uses the naive approach, converting to base 4 and counting the number of 1 and 2 digits.

## Reflection

The thing to note is that hgl isn't really well suited for arithmetic heavy challenges (yet).

• There should be a function that combines fl and nn to get natural numbers satisfying a property.
• I keep making variants of l2, but I feel like l2 eq would be a good shortcut to have. The precedence on (*=*) is incompatible with (<). I should change it to be anything else.
• bs should be made to handle negative bases as well. I do believe there is a way to do this question in terms of negative bases, although I haven't really investigated since I can't do negative bases in hgl.
• rF(-)0, or "alternating sum", could probably be added.

# Python 3, 107 bytes

f=lambda n,i=0,l=[]:i-1if n==sum(l)else f(n,i+1,l+[a(i,2)==a(i,3)])
a=lambda n,s:sum(map(int,bin(n)[s::2]))


Try it online!

Uses the same logic as other solutions to sum odd and even bits. Generates a list of T/F values corresponding to the index value's phat-ness, and returns the index found when the sum of the T/F list is equal to n.

# 05AB1E, 7 bytes

∞<ʒb®ö_


Outputs a lazy infinite list.

Try it online.

Or:

[Nb®ö_–


Infinitely outputs all results on a separated newline to STDOUT.

Try it online.

Both are a port of @JonathanAllan's Jelly answer.

Explanation:

∞        # Push an infinite positive list: [1,2,3,...]
<       # Make it non-negative by decreasing each by 1: [0,1,2,...]
ʒ      # Filter this list by:
b     #  Convert from a base-10 integer to a binary string
®ö   #  Convert from a base-(-1) integer to a base-10 integer
_  #  Check that this is equal to 0
# (after which the lazy finite list is output implicitly as result)

[        # Start an infinite loop:
N       #  Push the current 0-based loop index
b®ö_   #  Same as above
–  #  If this is truthy: output the 0-based loop index with trailing newline


# Scala 3, 212 bytes

A port of @Arnold Palmer's Python answer in Scala.

Golfed version. Attempt This Online!

def f(n:Int,i:Int=0,l:Seq[Boolean]=Seq()):Int={if(n==l.count(identity))i-1 else f(n,i+1,l:+(a(i,2)==a(i,3)))}
def a(n:Int,s:Int)=n.toBinaryString.zipWithIndex.collect{case(c,i) if(i+s)%2==0=>c.toString.toInt}.sum


Ungolfed version. Attempt This Online!

object Main {
def main(args: Array[String]): Unit = {
for (i <- 1 to 16) {
println(s"$i:${f(i)}")
}
}

def f(n: Int, i: Int = 0, l: List[Boolean] = List()): Int = {
if (n == l.count(identity)) i - 1
else f(n, i + 1, l :+ (a(i, 2) == a(i, 3)))
}

def a(n: Int, s: Int): Int = {
val binaryString = n.toBinaryString
// Create a sequence from the binary string starting at index s and taking every second character, then map to Int and sum
binaryString.zipWithIndex.collect {
case (char, index) if (index + s) % 2 == 0 => char.toString.toInt
}.sum
}
}


# C (gcc), 107 92 bytes

-15 thanks to @ceilingcat for much smarter for loop utilization

b[2],i,c,k;f(n){for(k=c=0;k^n;k+=*b==b[1],c++)for(*b=b[1]=i=0;i<32;)b[i%2]+=c>>i++&1;n=c-1;}


Try it online!

Uses a 2 element array to count odd and even bits, incrementing a count if equal, and returns the last value checked when the count was incremented.