A291711 -id:A291711 - cp's OEIS frontend

A381579 The Chung-Graham representation of n: representation of n in base of even-indexed Fibonacci numbers.

0, 1, 2, 10, 11, 12, 20, 21, 100, 101, 102, 110, 111, 112, 120, 121, 200, 201, 202, 210, 211, 1000, 1001, 1002, 1010, 1011, 1012, 1020, 1021, 1100, 1101, 1102, 1110, 1111, 1112, 1120, 1121, 1200, 1201, 1202, 1210, 1211, 2000, 2001, 2002, 2010, 2011, 2012, 2020, 2021

Offset: 0

Amiram Eldar, Feb 28 2025

Chung and Graham (1981, 1984) proved that every nonnegative integer n can be uniquely represented as a sum n = Sum_{i>=1} d_i * Fibonacci(2*i), where each d_i is in {0, 1, 2}, and if d_i = d_j = 2 with i < j, then for some k, i < k < j, we have d_k = 0.

			a(3) = 10 since 3 = 1 * 3 + 0 * 1 = 3 * Fibonacci(2*2) + 0 * Fibonacci(2*1).
a(10) = 102 since 10 = 1 * 8 + 0 * 3 + 2 * 1 = 1 * Fibonacci(2*3) + 0 * Fibonacci(2*2) + 2 * Fibonacci(2*1).

Amiram Eldar, Table of n, a(n) for n = 0..10000
Rob Burns, Chung-Graham and Zeckendorf representations, arXiv:2502.18870 [math.NT], 2025.
Hung Viet Chu, Aney Manish Kanji, and Zachary Louis Vasseur, Fixed-Term Decompositions Using Even-Indexed Fibonacci Numbers, arXiv:2501.03231 [math.GM], 2024.
F. R. K. Chung and R. L. Graham, On irregularities of distribution of real sequences, Proc. Nat. Acad. Sci. USA, Vol. 78, No. 7 (1981), p. 4001.
F. R. K. Chung and R. L. Graham, On irregularities of distribution, Finite and Infinite Sets, Colloquia Mathematica Societatis János Bolyai, Vol. 37, North-Holland, 1984, pp. 181-222; alternative link.

Cf. A000045, A001906, A291711 (sum of digits).

Similar sequences: A014417, A104326.

f[n_] := f[n] = Fibonacci[2*n]; a[n_] := Module[{s = 0, m = n, k}, While[m > 0, k = 1; While[m > f[k], k++]; If[m < f[k], k--]; If[m >= 2*f[k], s += 2*10^(k-1); m -= 2*f[k], s += 10^(k-1); m -= f[k]]]; s]; Array[a, 50, 0]

m = 20; fvec = vector(m, i, fibonacci(2*i)); f(n) = if(n <= m, fvec[n], fibonacci(2*n));
a(n) = {my(s = 0, m = n, k); while(m > 0, k = 1; while(m > f(k), k++); if(m < f(k), k--); if(m >= 2*f(k), s += 2*10^(k-1); m -= 2*f(k), s += 10^(k-1); m -= f(k))); s;}

A381581 Numbers divisible by the sum of the digits in their Chung-Graham representation (A381579).

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 12, 16, 20, 21, 22, 24, 27, 28, 30, 40, 42, 44, 45, 48, 55, 56, 57, 58, 60, 66, 70, 72, 75, 76, 80, 84, 90, 92, 95, 96, 100, 102, 110, 111, 112, 115, 116, 120, 132, 135, 138, 140, 144, 150, 152, 153, 156, 168, 170, 175, 176, 180, 186, 190, 195, 198

Offset: 1

Amiram Eldar, Feb 28 2025

Numbers k such that A291711(k) divides k.
Analogous to Niven numbers (A005349) with the Chung-Graham representation (A381579) instead of the decimal representation.
A001906(k) = Fibonacci(2*k) is a term for all k >= 1.
If k is not divisible by 3 (A001651), then Fibonacci(2*k) + 1 is a term.

			4 is a term since A291711(4) = 1 divides 4.
6 is a term since A291711(6) = 2 divides 6.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000045, A001651, A001906, A291711.
Subsequences: A381582, A381583, A381584, A381585.
Similar sequences: A005349, A049445, A064150, A328208, A328212.

f[n_] := f[n] = Fibonacci[2*n]; q[n_] := Module[{s = 0, m = n, k}, While[m > 0, k = 1; While[m > f[k], k++]; If[m < f[k], k--]; If[m >= 2*f[k], s += 2; m -= 2*f[k], s++; m -= f[k]]]; Divisible[n, s]]; Select[Range[200], q]

mx = 20; fvec = vector(mx, i, fibonacci(2*i)); f(n) = if(n <= mx, fvec[n], fibonacci(2*n));
isok(n) = {my(s = 0, m = n, k); while(m > 0, k = 1; while(m > f(k), k++); if(m < f(k), k--); if(m >= 2*f(k), s += 2; m -= 2*f(k), s++; m -= f(k))); !(n % s);}

A381582 Numbers k such that k and k+1 are both terms in A381581.

Original entry on oeis.org

1, 2, 3, 20, 21, 27, 44, 55, 56, 57, 75, 95, 110, 111, 115, 152, 175, 207, 264, 287, 291, 304, 305, 344, 364, 365, 377, 380, 395, 398, 399, 404, 425, 435, 455, 534, 584, 605, 815, 846, 847, 864, 888, 930, 987, 992, 1004, 1011, 1024, 1025, 1064, 1084, 1085, 1145, 1182

Offset: 1

Amiram Eldar, Feb 28 2025

If k is not divisible by 3 (A001651), then A001906(k) = Fibonacci(2*k) is a term.

			1 is a term since A291711(1) = 1 divides 1 and A291711(2) = 2 divides 2.
20 is a term since A291711(20) = 4 divides 20 and A291711(21) = 1 divides 21.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000045, A001906, A001651, A291711.
Subsequence of A381581.
Subsequences: A381583, A381584, A381585.
Similar sequences: A328209, A330927, A330931, A351720.

f[n_] := f[n] = Fibonacci[2*n]; q[n_] := q[n] = Module[{s = 0, m = n, k}, While[m > 0, k = 1; While[m > f[k], k++]; If[m < f[k], k--]; If[m >= 2*f[k], s += 2; m -= 2*f[k], s++; m -= f[k]]]; Divisible[n, s]]; Select[Range[1200], q[#] && q[#+1] &]

mx = 20; fvec = vector(mx, i, fibonacci(2*i)); f(n) = if(n <= mx, fvec[n], fibonacci(2*n));
is1(n) = {my(s = 0, m = n, k); while(m > 0, k = 1; while(m > f(k), k++); if(m < f(k), k--); if(m >= 2*f(k), s += 2; m -= 2*f(k), s++; m -= f(k))); !(n % s);}
list(lim) = {my(q1 = is1(1), q2); for(k = 2, lim, q2 = is1(k); if(q1 && q2, print1(k-1, ", ")); q1 = q2);}

A381583 Starts of runs of 3 consecutive integers that are all terms in A381581.

Original entry on oeis.org

1, 2, 20, 55, 56, 110, 304, 364, 398, 846, 1024, 1084, 1744, 1854, 2044, 2104, 2105, 2527, 2824, 2862, 3870, 4374, 5222, 5223, 5243, 5718, 5928, 6488, 6784, 6844, 6894, 6978, 7142, 7924, 10590, 11240, 11889, 11975, 12248, 14284, 14915, 16638, 17710, 17714, 17824

Offset: 1

Amiram Eldar, Feb 28 2025

If k is congruent to 1 or 5 mod 12 (A087445), then A001906(k) = Fibonacci(2*k) is a term.

			1 is a term since A291711(1) = 1 divides 1, A291711(2) = 2 divides 2, and A291711(3) = 1 divides 3.
20 is a term since A291711(20) = 4 divides 20, A291711(21) = 1 divides 21, and A291711(22) = 2 divides 22.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000045, A001906, A087445, A291711.
Subsequence of A381581 and A381582.
Subsequences: A381584, A381585.
Similar sequences: A154701, A328210, A330932, A351721.

f[n_] := f[n] = Fibonacci[2*n]; q[n_] := q[n] = Module[{s = 0, m = n, k}, While[m > 0, k = 1; While[m > f[k], k++]; If[m < f[k], k--]; If[m >= 2*f[k], s += 2; m -= 2*f[k], s++; m -= f[k]]]; Divisible[n, s]]; seq[count_, nConsec_] := Module[{cn = q /@ Range[nConsec], s = {}, c = 0, k = nConsec + 1}, While[c < count, If[And @@ cn, c++; AppendTo[s, k - nConsec]]; cn = Join[Rest[cn], {q[k]}]; k++]; s]; seq[45, 3]

mx = 20; fvec = vector(mx, i, fibonacci(2*i)); f(n) = if(n <= mx, fvec[n], fibonacci(2*n));
is1(n) = {my(s = 0, m = n, k); while(m > 0, k = 1; while(m > f(k), k++); if(m < f(k), k--); if(m >= 2*f(k), s += 2; m -= 2*f(k), s++; m -= f(k))); !(n % s);}
list(lim) = {my(q1 = is1(1), q2 = is1(2), q3); for(k = 3, lim, q3 = is1(k); if(q1 && q2 && q3, print1(k-2, ", ")); q1 = q2; q2 = q3);}

A381584 Starts of runs of 4 consecutive integers that are all terms in A381581.

Original entry on oeis.org

1, 55, 2104, 5222, 24784, 63510, 64264, 69487, 95463, 121393, 184327, 327303, 374589, 463110, 468168, 561069, 572550, 596868, 671407, 740310, 759030, 819948, 902670, 956680, 1023009, 1036230, 1065030, 1259817, 1274910, 1359552, 1683154, 1714470, 1731750, 2182023

Offset: 1

Amiram Eldar, Feb 28 2025

If k is congruent to 1 or 5 mod 12 (A087445), then A001906(k) = Fibonacci(2*k) is a term.

			1 is a term since A291711(1) = 1 divides 1, A291711(2) = 2 divides 2, A291711(3) = 1 divides 3, and A291711(4) = 2 divides 4.
55 is a term since A291711(55) = 1 divides 55, A291711(56) = 2 divides 56, A291711(57) = 3 divides 57, and A291711(58) = 2 divides 58.

Amiram Eldar, Table of n, a(n) for n = 1..1000

Cf. A000045, A001906, A087445, A291711.
Subsequence of A381581, A381582 and A381583.
A381585 is a subsequence.
Similar sequences: A141769, A328211, A328215, A330933.

f[n_] := f[n] = Fibonacci[2*n]; q[n_] := q[n] = Module[{s = 0, m = n, k}, While[m > 0, k = 1; While[m > f[k], k++]; If[m < f[k], k--]; If[m >= 2*f[k], s += 2; m -= 2*f[k], s++; m -= f[k]]]; Divisible[n, s]]; seq[count_, nConsec_] := Module[{cn = q /@ Range[nConsec], s = {}, c = 0, k = nConsec + 1}, While[c < count, If[And @@ cn, c++; AppendTo[s, k - nConsec]]; cn = Join[Rest[cn], {q[k]}]; k++]; s]; seq[12, 4]

mx = 20; fvec = vector(mx, i, fibonacci(2*i)); f(n) = if(n <= mx, fvec[n], fibonacci(2*n));
is1(n) = {my(s = 0, m = n, k); while(m > 0, k = 1; while(m > f(k), k++); if(m < f(k), k--); if(m >= 2*f(k), s += 2; m -= 2*f(k), s++; m -= f(k))); !(n % s);}
list(lim) = {my(q1 = is1(1), q2 = is1(2), q3 = is1(3), q4); for(k = 4, lim, q4 = is1(k); if(q1 && q2 && q3 && q4, print1(k-3, ", ")); q1 = q2; q2 = q3; q3 = q4);}

A381585 Starts of runs of 5 consecutive integers that are all terms in A381581.

Original entry on oeis.org

57744971, 159104411, 203738652, 212548572, 260463851, 361823291, 413644572, 431577521, 440353328, 520800012, 717222337, 726300972, 779825648, 843559091, 913313321, 945016812, 986681527, 1091786528, 1116032201, 1185786431, 1318751081, 1347208812, 1360423692, 1418212627

Offset: 1

Amiram Eldar, Feb 28 2025

Are there 6 consecutive integers that are all terms in A381581?

			57744971 is a term since A291711(57744971) = 17 divides 57744971, A291711(57744972) = 18 divides 57744972, A291711(57744973) = 13 divides 57744973, A291711(57744974) = 14 divides 57744974, and A291711(57744975) = 15 divides 57744975.

Amiram Eldar, Table of n, a(n) for n = 1..130

Cf. A291711.
Subsequence of A381581, A381582, A381583 and A381584.
Similar sequences: A330928, A334373, A364220, A364383.

mx = 20; fvec = vector(mx, i, fibonacci(2*i)); f(n) = if(n <= mx, fvec[n], fibonacci(2*n));
is1(n) = {my(s = 0, m = n, k); while(m > 0, k = 1; while(m > f(k), k++); if(m < f(k), k--); if(m >= 2*f(k), s += 2; m -= 2*f(k), s++; m -= f(k))); !(n % s);}
list(lim) = {my(q1 = is1(1), q2 = is1(2), q3 = is1(3), q4 = is1(4), s5); for(k = 5, lim, q5 = is1(k); if(q1 && q2 && q3 && q4 && q5, print1(k-4, ", ")); q1 = q2; q2 = q3; q3 = q4; q4 = q5);}

cp's OEIS Frontend

A381579 The Chung-Graham representation of n: representation of n in base of even-indexed Fibonacci numbers.

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A381581 Numbers divisible by the sum of the digits in their Chung-Graham representation (A381579).

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A381582 Numbers k such that k and k+1 are both terms in A381581.

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A381583 Starts of runs of 3 consecutive integers that are all terms in A381581.

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A381584 Starts of runs of 4 consecutive integers that are all terms in A381581.

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A381585 Starts of runs of 5 consecutive integers that are all terms in A381581.

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