Ctibor O. Zizka - cp's OEIS frontend

A387123 Numbers k such that Sum_{i=1..r} (k-i) and Sum_{i=1..r} (k+i) are both triangular for some r with 1 <= r < k.

2, 6, 9, 21, 24, 38, 50, 53, 65, 77, 90, 96, 104, 133, 147, 195, 201, 224, 247, 286, 324, 377, 450, 483, 553, 588, 605, 614, 713, 792, 901, 1014, 1029, 1043, 1066, 1074, 1155, 1274, 1349, 1575, 1784, 1885, 1920, 2034, 2057, 2109, 2279, 2312, 2342, 2622

Offset: 1

Ctibor O. Zizka, Aug 17 2025

For m >= 1, if k = m*(m+1)^2/2 then r = m, thus A006002 is a subsequence. For k >= 286 from A101265 or A101879, r = k-1.

			For k = 6: the least r = 5, T_i = 1 + 2 + 3 + 4 + 5 = 15, T_j = 7 + 8 + 9 + 10 + 11 = 45, both T_i and T_j are triangular numbers, thus k = 6 is a term.

Cf. A000217, A006002, A101265, A101879.

triQ[n_] := IntegerQ[Sqrt[8*n + 1]]; q[k_] := Module[{r = 1, s1 = 0, s2 = 0}, While[s1 += k - r; s2 += k + r; r < k && (! triQ[s1] || ! triQ[s2]), r++]; 1 <= r < k]; Select[Range[3000], q] (* Amiram Eldar, Aug 17 2025 *)

isok(k) = my(sm=0, sp=0); for (r=1, k-1, sm+=k-r; sp+=k+r; if (ispolygonal(sm, 3) && ispolygonal(sp, 3), return(r));); \\ Michel Marcus, Aug 17 2025

from itertools import count, islice
from sympy.ntheory.primetest import is_square
def A387123_gen(startvalue=1): # generator of terms >= startvalue
    for k in count(max(startvalue,1)):
        if any(is_square(((k*r<<1)-r*(r+1)<<2)+1) and is_square(((k*r<<1)+r*(r+1)<<2)+1) for r in range(1,k)):
            yield k
A387123_list = list(islice(A387123_gen(),50)) # Chai Wah Wu, Aug 21 2025

A386971 Numbers k >= 1 such that w(k-r) + ... + w(k-1) = w(k+1) + ... + w(k+r) for some r >= 1 where w(i) is the binary weight of i (A000120).

Original entry on oeis.org

3, 4, 7, 8, 11, 12, 14, 15, 16, 17, 19, 20, 23, 24, 27, 28, 29, 31, 32, 34, 35, 36, 39, 40, 43, 44, 46, 47, 48, 49, 51, 52, 55, 56, 58, 59, 60, 63, 64, 67, 68, 69, 71, 72, 75, 76, 78, 79, 80, 81, 83, 84, 87, 88, 91, 92, 93, 95, 96, 98, 99, 100, 103, 104, 107

Offset: 1

Ctibor O. Zizka, Aug 11 2025

r = 1 for k from A047457, k = 8*x - 5 or k = 8*x - 4, x >= 1.

r = 2 for k = 32*x - 18, (2* 7th row of A097586) or k = 32*x - 15, (10th row of A097586), x >= 1.

			For k = 7: A000120(4) + A000120(5) + A000120(6) = A000120(8) + A000120(9) + A000120(10), thus 7 is a term.

Ctibor O. Zizka, A386971 plot of k vs. r

Cf. A000120, A047457, A097586.

q[k_] := Module[{s = 0, r = 1}, While[r < k && (r == 1 || s != 0), s += (DigitSum[k-r, 2] - DigitSum[k+r, 2]); r++]; 1 < r <= k && s ==0]; Select[Range[120], q] (* Amiram Eldar, Aug 12 2025 *)

A385905 Numbers k >= 1 such that digsum(k-r) + ... + digsum(k-1) = digsum(k+1) + ... + digsum(k+r) for some r >= 1 where digsum(i) is the digital sum of i (A007953).

Original entry on oeis.org

9, 10, 19, 20, 29, 30, 39, 40, 49, 50, 59, 60, 69, 70, 77, 79, 80, 85, 86, 89, 90, 91, 92, 99, 100, 107, 108, 109, 110, 113, 114, 119, 120, 122, 129, 130, 139, 140, 149, 150, 159, 160, 169, 170, 177, 179, 180, 185, 186, 189, 190, 191, 192, 197, 199, 200, 202

Offset: 1

Ctibor O. Zizka, Aug 12 2025

Empirical observation: k != A214678(n).

			For k = 9: the least r = 8, A007953(1) + ... + A007953(8) = A007953(10) + ... + A007953(17), thus k = 9 is a term.

Ctibor O. Zizka, A385905 plot of k vs. r

Cf. A007953, A214678.

q[k_] := Module[{s = 0, r = 1}, While[r < k && (r == 1 || s != 0), s += (DigitSum[k-r] - DigitSum[k+r]); r++];1 < r <= k && s ==0]; Select[Range[202], q] (* Amiram Eldar, Aug 12 2025 *)

A386987 For n >= 2, a(n) is the least r >= 1 such that T(n - r) + ... + T(n - 1) = T(n + 1) + ... + T(n + r) where T(i) is A010060(i).

Original entry on oeis.org

2, 1, 1, 2, 4, 3, 3, 4, 2, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 4, 3, 3, 4, 2, 1, 1, 2, 4, 3, 3, 4, 2, 1, 1, 2, 4, 3, 3, 4, 2, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 4, 3, 3, 4, 2, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 4, 3, 3, 4, 2, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 4, 3, 3

Offset: 2

Ctibor O. Zizka, Aug 12 2025

nonn

a(n) is from {1, 2, 3, 4}.

			For n = 6: T(6 - r) + ... + T(5) = T(7) + ... + T(6 + r) is true for the least r = 4  because A010060(2) + A010060(3) + A010060(4) + A010060(5) = A010060(7) + A010060(8) + A010060(9) + A010060(10), thus a(6) = 4.

Cf. A010060, A056196, A079523, A081706, A131323, A225822, A249034.

a[n_] := Module[{s = 0, r = 1}, While[r <= n && (r == 1 || s != 0), s += (ThueMorse[n - r] - ThueMorse[n + r]); r++]; r-1]; Array[a, 100, 2] (* Amiram Eldar, Aug 12 2025 *)

a(A081706(n) + 1) = 1.
a(2*A079523(n)) = 2.
a(A249034(n))= 2.
a(A225822(n)) = 3.
a(A056196(n)) = 3.
a(2*A131323(n)) = 4.
a(2*A249034(n) - 1) = 4.

A386743 For n >= 0, a(n) is the least Fibonacci number F(i) such that F(i) = (2n + 1)(n + k) for some k >= 0.

Original entry on oeis.org

0, 3, 55, 21, 144, 55, 377, 6765, 2584, 2584, 987, 46368, 75025, 14930352, 317811, 832040, 6765, 102334155, 4181, 317811, 6765, 701408733, 1548008755920, 2178309, 225851433717, 14930352, 196418, 6765, 14930352, 591286729879, 832040, 46368, 9227465, 72723460248141, 46368

Offset: 0

Ctibor O. Zizka, Aug 01 2025

nonn

a(n) >= n*(2*n + 1).

Empirical observation: For F(i), i is from {1/3, 1/2, 2/3, 1, 3/2, 2}*A001177(2*n + 1).

			For n = 3: F(i) = 21 + 7*k is true for k = 0, F(8) = 21, thus a(3) = 21.

Cf. A000045, A001177.

a[n_]:=Module[{i=0},While[!Divisible[Fibonacci[i],2n+1] || Fibonacci[i]/(2n+1) < n, i++]; Fibonacci[i]]; Array[a,35,0] (* Stefano Spezia, Aug 04 2025 *)

a(n) = my(i=0, f=fibonacci(0)); while((f % (2*n+1)) || (f/(2*n+1) < n), i++; f=fibonacci(i)); f; \\ Michel Marcus, Aug 01 2025

def A386743(n):
    a, b, m, k = 0, 1, n*((n<<1)|1), (n<<1)|1
    while aChai Wah Wu, Aug 11 2025

More terms from Michel Marcus, Aug 01 2025

A385655 Numbers k >= 0 such that k = digsum(k) + digsum(k+1) + ... + digsum(k+r) for some r >= 0 where digsum(i) is the digital sum of i (A007953).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 17, 19, 20, 22, 28, 29, 34, 37, 38, 46, 47, 51, 56, 65, 70, 79, 88, 91, 94, 95, 103, 105, 106, 112, 114, 118, 121, 123, 130, 132, 137, 141, 150, 157, 166, 175, 178, 184, 192, 200, 203, 205, 220, 222, 235, 241, 246, 260

Offset: 1

Ctibor O. Zizka, Aug 03 2025

			For k = 10: 10 = A007953(10) + A007953(11) + A007953(12) + A007953(13) = 10, thus 10 is a term.

Ctibor O. Zizka, Relationship between k and r

Cf. A007953.

q[k_] := Module[{s = 0, m = k}, While[s < k, s += DigitSum[m]; m++]; s == k]; Select[Range[0, 300], q] (* Amiram Eldar, Aug 03 2025 *)

isok(k) = my(s=sumdigits(k), t=k+1); while (s < k, s+=sumdigits(t); t++); s==k; \\ Michel Marcus, Aug 04 2025

from itertools import count, islice
def A385655_gen(startvalue=0): # generator of terms >= startvalue
    for k in count(max(startvalue,0)):
        s, r = 0, k
        while sA385655_list = list(islice(A385655_gen(),40)) # Chai Wah Wu, Aug 09 2025

A386277 For n >= 0, a(n) is the least Fibonacci number F_i (A000045) such that for some k >= 0, F_i = (n + 1)(n + 2k)/2, or a(n) = -1 if no such k exists.

Original entry on oeis.org

0, 1, 3, 34, 55, 21, 21, -1, 144, 55, 55, -1, 377, 987, 6765, 2584, 2584, -1, 2584, 610, 987, 6765, 46368, -1, 75025, 377, 14930352, -1, 317811, 6765, 832040, 14930352, 6765, -1, 102334155, -1, 4181, -1, 317811, -1, 6765, 987, 701408733, -1, 1548008755920, -1, 2178309, -1, 225851433717

Offset: 0

Ctibor O. Zizka, Jul 17 2025

sign

n even: F_i exists for all even n. Proof: For n even we have (n*(n + 1)/2) mod (n + 1) = 0. The residue 0 is among the residues of the Pisano period of (Fibonacci sequence mod (n + 1)) for all  n.
n odd: F_i exists only if the residue (n*(n + 1)/2) mod (n + 1) = (n + 1)/2 is among the residues of the Pisano period of (Fibonacci sequence mod (n + 1)).
Computational result: it looks like a(2*A367420(n) - 1) = -1, checked for n from [1, 2000].
The lower bound for nonnegative a(n) is n*(n + 1)/2.
Empirical observation: for a(n) = F_i, i = t*A001177(n + 1), t is from {1/3, 1/2, 2/3, 1, 3/2, 2}.

			For n = 3: (3 + 1)*(3 + 2*k)/2 = 6 + 4*k is a Fibonacci number at first for k = 7. Thus a(3) = 34.
For n = 7: (7 + 1)*(7 + 2*k)/2 = 28 + 8*k. (Fib_i congruent to 4 mod 8) has no solution because the Fibonacci sequence modulo 8 has a period of 12: (1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0), the residues are {0, 1, 2, 3, 5, 7} and 4 is not among them. Thus a(7) = -1.
For n = 11: (11 + 1)*(11 + 2*k)/2 = 66 + 12*k . (Fib_i congruent to 6 mod 12) has no solution because the Fibonacci sequence modulo 12 has a period of 24: (1, 1, 2, 3, 5, 8, 1, 9, 10, 7, 5, 0, 5, 5, 10, 3, 1, 4, 5, 9, 2, 11, 1, 0), the residues are {0, 1, 2, 3, 4, 5, 7, 8, 9, 10, 11} and 6 is not among them. Thus a(11) = -1.

Michel Marcus, Table of n, a(n) for n = 0..3643
Ctibor O. Zizka, Ratio of i/Z(n+1) vs. n for n = 1 to 500. Z(n+1) = A001177(n+1)

Cf. A000045, A000217, A001175, A001177, A161553, A367420.

isfib(n) = my(k=n^2); k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8));
row(n) = {my(L=List([0]), X=Mod([1, 1; 1, 0], n), I=Mod([1, 0; 0, 1], n), M=X); while(M<>I, M*=X; listput(L, lift(M[2, 2]))); Vec(L);} \\ A161553
a(n) = my(p=(n + 1)*(n + 2*k)/2, x=polcoef(p, 1, 'k), y = polcoef(p, 0, 'k), v=row(x)); if (!vecsearch(Set(v),y % x), return(-1)); my(j=0); while ((denominator(jj=((2*fibonacci(j) - n*(n+1))/(2*(n+1)))) != 1) || (numerator(jj)<0), j++); fibonacci(j); \\ Michel Marcus, Jul 19 2025

Corrected and extended by Michel Marcus, Jul 19 2025

A386546 Numbers k >= 1 such that k = d(k) + d(k+1) + ... + d(k+r) for some r >= 0 where d(i) is the number of divisors of i (A000005).

Original entry on oeis.org

1, 2, 6, 9, 12, 18, 21, 26, 28, 31, 43, 49, 52, 54, 73, 79, 91, 93, 95, 99, 102, 109, 111, 121, 122, 133, 153, 159, 175, 179, 185, 193, 197, 211, 215, 227, 231, 239, 241, 243, 271, 279, 286, 291, 295, 299, 301, 305, 309, 311, 313, 318, 324, 329, 339, 345

Offset: 1

Ctibor O. Zizka, Jul 25 2025

nonn

r is from {0,0,1,2,2,3,4,5,5,6,8,9,9,13,13,13,16,15,15,...}.

			For k = 2: 2 = A000005(2) = 2, thus 2 is a term.
For k = 6: 6 = A000005(6) + A000005(7) = 4 + 2 = 6, thus 6 is a term.

Cf. A000005.

q[k_] := Module[{s = 0, m = k}, While[s < k, s += DivisorSigma[0, m]; m++]; s == k]; Select[Range[350], q] (* Amiram Eldar, Jul 25 2025 *)

A385738 For n >= 1, a(n) is the least k such that the Sum_{i=0..(n-1)} (k+i)/A000005(k+i) is an integer or a(n) = -1 if no such k exists.

Original entry on oeis.org

1, 1, 6, 6, 8, 5, 6, 23, 5, 22, 50, 26, 28, 65, 119, 145, 26, 349, 282, 375, 280, 404, 278, 369, 279, 370, 277, 276, 369, 378, 389, 378, 389, 15, 389, 13, 12, 210, 10, 9, 8, 210, 6, 212, 421, 209, 419, 3, 2, 1, 378, 419, 421, 418, 418, 1026, 373, 105, 104

Offset: 1

Ctibor O. Zizka, Jul 08 2025

nonn

			For n = 4: Sum_{i=0..3} (k+i)/A000005(k+i) is an integer for the least k = 6 because 6/A000005(6) + 7/A000005(7) + 8/A000005(8) + 9/A000005(9) = 10.

Cf. A000005.

seq[max_] := Module[{v = Table[n/DivisorSigma[0, n], {n, 1, max}], k = 1, s = {}, i}, While[NumberQ[i = FirstPosition[Plus @@@ Partition[v, k, 1], ?IntegerQ][[1]]], AppendTo[s, i]; k++]; s]; seq[1200] (* _Amiram Eldar, Jul 08 2025 *)

a(n) = my(k=1); while(denominator(sum(i=0, n-1, (k+i)/numdiv(k+i))) != 1, k++); k; \\ Michel Marcus, Jul 08 2025

A385720 Numbers k >= 1 such that k/A000005(k) + (k+1)/A000005(k+1) is an integer.

Original entry on oeis.org

1, 5, 6, 8, 10, 13, 22, 37, 45, 46, 58, 61, 62, 69, 73, 74, 77, 82, 89, 106, 114, 117, 126, 146, 149, 150, 154, 157, 166, 167, 178, 186, 193, 197, 198, 206, 221, 226, 233, 237, 258, 261, 262, 263, 266, 277, 278, 279, 280, 290, 293, 306, 309, 311, 312, 313

Offset: 1

Ctibor O. Zizka, Jul 07 2025

nonn

k/A000005(k) + (k+1)/A000005(k+1) = (3*k + 1)/4 for k >= 5 from A256072.
k/A000005(k) + (k+1)/A000005(k+1) = (3*k + 2)/4 for k >= 6 from A077065.
k/A000005(k) + (k+1)/A000005(k+1) =< (3*k + 2)/4 for k >= 5.

			For k = 6: 6/A000005(6) + 7/A000005(7) = 6/4 + 7/2 = 5, thus k = 6 is a term.

Cf. A000005, A005384, A005385, A077065, A256072.

Position[Plus @@@ Partition[Table[n/DivisorSigma[0, n], {n, 1, 320}], 2, 1], ?IntegerQ] // Flatten (* _Amiram Eldar, Jul 08 2025 *)

isok(k) = denominator(k/numdiv(k) + (k+1)/numdiv(k+1)) == 1; \\ Michel Marcus, Jul 08 2025

from itertools import count, islice
from sympy import divisor_count
def A385720_gen(startvalue=1): # generator of terms >= startvalue
    m = max(startvalue,1)
    a, b = divisor_count(m), divisor_count(m+1)
    for k in count(m):
        if not (k*b+(k+1)*a)%(a*b):
            yield k
        a, b = b, divisor_count(k+2)
A385720_list = list(islice(A385720_gen(),30)) # Chai Wah Wu, Jul 13 2025

cp's OEIS Frontend

User: Ctibor O. Zizka

A387123 Numbers k such that Sum_{i=1..r} (k-i) and Sum_{i=1..r} (k+i) are both triangular for some r with 1 <= r < k.

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A386971 Numbers k >= 1 such that w(k-r) + ... + w(k-1) = w(k+1) + ... + w(k+r) for some r >= 1 where w(i) is the binary weight of i (A000120).

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A385905 Numbers k >= 1 such that digsum(k-r) + ... + digsum(k-1) = digsum(k+1) + ... + digsum(k+r) for some r >= 1 where digsum(i) is the digital sum of i (A007953).

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A386987 For n >= 2, a(n) is the least r >= 1 such that T(n - r) + ... + T(n - 1) = T(n + 1) + ... + T(n + r) where T(i) is A010060(i).

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A386743 For n >= 0, a(n) is the least Fibonacci number F(i) such that F(i) = (2*n + 1)*(n + k) for some k >= 0.

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A385655 Numbers k >= 0 such that k = digsum(k) + digsum(k+1) + ... + digsum(k+r) for some r >= 0 where digsum(i) is the digital sum of i (A007953).

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A386277 For n >= 0, a(n) is the least Fibonacci number F_i (A000045) such that for some k >= 0, F_i = (n + 1)*(n + 2*k)/2, or a(n) = -1 if no such k exists.

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A386546 Numbers k >= 1 such that k = d(k) + d(k+1) + ... + d(k+r) for some r >= 0 where d(i) is the number of divisors of i (A000005).

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A386743 For n >= 0, a(n) is the least Fibonacci number F(i) such that F(i) = (2n + 1)(n + k) for some k >= 0.

A386277 For n >= 0, a(n) is the least Fibonacci number F_i (A000045) such that for some k >= 0, F_i = (n + 1)(n + 2k)/2, or a(n) = -1 if no such k exists.