A000217 -id:A000217 - cp's OEIS frontend

A137281 Numbers k such that T(k) is not divisible by T(i), 1 < i < k, where T(k) = k-th triangular number A000217(k).

2, 4, 7, 10, 13, 16, 22, 25, 28, 31, 34, 37, 43, 46, 49, 52, 58, 61, 67, 70, 73, 76, 82, 85, 88, 94, 97, 103, 106, 118, 121, 127, 130, 133, 136, 142, 145, 148, 151, 157, 163, 166, 169, 172, 178, 187, 190, 193, 196, 202, 205, 208, 211, 214, 217, 226, 229, 232, 238, 241

Offset: 1

Zak Seidov, Mar 14 2008

nonn

All terms > 5 in A005383 are here. - Zak Seidov, Jun 20 2013
All terms except 2 are congruent to 1 (mod 3). This is required for 3 not to be a divisor of T(n). - Franklin T. Adams-Watters, Dec 10 2019
Conjecture: a(n) ~ C * n * log(n) for some constant C, in analogy with the prime number theorem (see A000040). - Harry Richman, Mar 05 2025

			T(4)=10 is not divisible by lesser T's 3, 6;
T(7)=28 is not divisible by lesser T's 3, 6, 10, 15, 21.

Zak Seidov, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)

Cf. A000217, A005383, A076982, A226863.

nn = 241; tri = Table[n*(n+1)/2, {n, nn}]; Select[Range[2, nn], ! MemberQ[Mod[tri[[#]], Take[tri, {2, # - 1}]], 0] &] (* T. D. Noe, Apr 12 2011 *)

n such that A076982(n) = 2. - T. D. Noe, Apr 12 2011

A000217(a(n)) = A226863(n). - Zak Seidov, Jun 20 2013

A138796 Least possible k > 0 with T(k) - T(j) = n, j > 0, where T(i) > 0 are the triangular numbers A000217.

Original entry on oeis.org

2, 3, 4, 3, 6, 4, 8, 4, 10, 6, 5, 7, 5, 6, 16, 9, 6, 10, 6, 8, 7, 12, 9, 7, 8, 7, 28, 15, 8, 16, 32, 8, 10, 8, 13, 19, 11, 9, 10, 21, 9, 22, 9, 10, 13, 24, 17, 10, 12, 11, 10, 27, 10, 13, 11, 12, 16, 30, 11, 31, 17, 11, 64, 11, 18, 34, 12, 14, 13, 36, 12, 37, 20, 12, 13, 12, 21, 40, 18

Offset: 2

Peter Pein (petsie(AT)dordos.net), Mar 30 2008

nonn

For T(k) see A138797, for j see A138798 and for T(j) see A138799.
The number of ways n can be written as difference of two triangular numbers is sequence A136107
Note that n = t(k)-t(j) implies 2n = (k-j)(k+j+1), where (k-j) and (k+j+1) are of opposite parity.  Let d be the odd element of { k-j, k+j+1 }. Then d is an odd divisor of n and k = ( d + 2n/d - 1 ) / 2. Therefore a(n) = ( min{ d + 2n/d } - 1 ) / 2 where d runs through all odd divisors of n, except perhaps (sqrt(8*n+1) +- 1)/2 which correspond to j=0. See PARI program. The restriction that j > 0 seems artificial. If it is removed we get A212652. - Max Alekseyev, Mar 31 2008

			a(30)=8, because 30 = T(30) - T(29) = T(11) - T(8) = T(9) - T(5) = T(8) - T(3) and 8 is the least index of the minuends.

Vincenzo Librandi, Table of n, a(n) for n = 2..1000
Peter Pein, Mathematica notebook containing a faster algorithm.

Cf. A000217, A109814, A118235, A136107, A138797, A138798, A138799, A212652.

T=#(#+1)/2&;Min[k/.{ToRules[Reduce[{T[k]-T[j]\[Equal]#,0

{ a(n) = local(m); m=2*n+1; fordiv(n/2^valuation(n,2),d,if((2*d+1)^2!=8*n+1&&(2*d-1)^2!=8*n+1,m=min(m,d+(2*n)\d))); (m-1)\2 }
vector(100,n,a(n)) \\ Max Alekseyev, Mar 31 2008

A181388 a(n) = Sum_{k=1..n} 2^T(k-1), where T = A000217 are the triangular numbers 0, 1, 3, 6, 10, ... . For n=0 we have the empty sum equal to 0.

Original entry on oeis.org

0, 1, 3, 11, 75, 1099, 33867, 2131019, 270566475, 68990043211, 35253362132043, 36064050381096011, 73823040345219302475, 302305277944002512979019, 2476182383848704552311227467, 40567295389687189552446813799499, 1329268563080305560093359507094144075

Offset: 0

Roman Pearce, Oct 17 2010

Original definition: The binary representation of each integer in the sequence consists of a single leading bit, followed by a string of n-1 zeros, followed by the previous integer. i.e. 3 = 2^1 + 1, 11 = 2^(2+1) + 3, 75 = 2^(3+2+1) + 11, and so on.

Numbers in this sequence may be used as a multiplier in hash functions to scatter and interleave bits.

Harvey P. Dale, Table of n, a(n) for n = 0..82

Cf. A000217, A006125.

f := proc(n) option remember; f(n-1) + 2^(ilog2(f(n-1))+ n - 1); end proc:
f(0) := 0:f(1):= 1:
seq(f(n),n=0..60); # updated by Robert Israel, Aug 28 2014

Join[{0},Accumulate[2^Accumulate[Range[0,15]]]] (* Harvey P. Dale, Mar 10 2016 *)

a(n)=sum(k=1,n,2^(k*(k-1)/2)) \\ M. F. Hasler, Aug 28 2014

a(n) = Sum_{k=1..n} A006125(k). - R. J. Mathar, Oct 18 2010

a(n) = a(n-1) + 2*(a(n-1) - a(n-2))^2/(a(n-2) - a(n-3)) for n >= 3. - Robert Israel, Aug 28 2014

Prefixed initial term a(0)=0 and simplified definition - M. F. Hasler, Aug 28 2014

A215795 Numbers n such that 2^n-1 is a triangular number (A000217).

Original entry on oeis.org

0, 1, 2, 4, 12

Offset: 1

V. Raman, Aug 23 2012

Aside from a(2), all terms are even. Probably complete; no more terms up to 10^6. - Charles R Greathouse IV, Sep 07 2012
This sequence maps to the Ramanujan-Nagell squares (8*(2^n - 1) + 1) and is therefore complete. - Raphie Frank, Sep 10 2012
Define equivalence classes on a specified real interval with respect to the symmetric transitive closure of R(x,y) = "x is an integer multiple of y". If any equivalence class is finite (the conditions for which are given in A328129), then a smallest equivalence class has cardinality 1, 2, 4 or 12. - Peter Munn, Jun 02 2021

Eric Weisstein's World of Mathematics, Ramanujan's Square Equation

Cf. A000217, A060728, A038198.
Cf. A076046 (triangular numbers of the form 2^n - 1).
Cf. A060728 (a(n) + 3).
Cf. A038198 (sqrt(8*(2^n - 1)+1)).
Cf. A215797 ((sqrt(8*(2^n - 1)+1) - 1)/2).
Cf. A328129.

Select[Range[0,15],OddQ[Sqrt[8(2^#-1)+1]]&] (* Harvey P. Dale, Dec 13 2024 *)

is(n)=issquare(8<Charles R Greathouse IV, Sep 07 2012

Four cross-references to the Ramanujan-Nagell problem added by Raphie Frank, Sep 10 2012

A222403 Triangle read by rows: left and right edges are A000217, interior entries are filled in using the Pascal triangle rule.

Original entry on oeis.org

0, 1, 1, 3, 2, 3, 6, 5, 5, 6, 10, 11, 10, 11, 10, 15, 21, 21, 21, 21, 15, 21, 36, 42, 42, 42, 36, 21, 28, 57, 78, 84, 84, 78, 57, 28, 36, 85, 135, 162, 168, 162, 135, 85, 36, 45, 121, 220, 297, 330, 330, 297, 220, 121, 45, 55, 166, 341, 517, 627, 660, 627, 517, 341, 166, 55

Offset: 0

N. J. A. Sloane, Feb 18 2013

In general, if the sequence defining the left and right edges is [a_0, a_1, ...], the row sums [s_0, s_1, ...] are given by s_0=a_0 and, for n>0,
s_n = 2a_n + Sum_{i=1..n-1} 2^(n-i) a_i.
Conversely, given the rows sums [s_0, s_1, ...], the edge sequence is [a_0, a_1, ...] where a_0=s_0 and, for n>0, a_n = (s_n - Sum_{i=1..n-1} s_i)/2.

			Triangle begins:
0
1, 1
3, 2, 3
6, 5, 5, 6
10, 11, 10, 11, 10
15, 21, 21, 21, 21, 15
21, 36, 42, 42, 42, 36, 21
28, 57, 78, 84, 84, 78, 57, 28
...

Robert Israel, Table of n, a(n) for n = 0..10010

Other triangles of this type: A007318, A051666, A134634, A222404, A222405.
Cf. A000217.
Row sums are A005803.

d:=[seq(n*(n+1)/2,n=0..14)];
f:=proc(d) local T,M,n,i;
M:=nops(d);
T:=Array(0..M-1,0..M-1);
for n from 0 to M-1 do T[n,0]:=d[n+1]; T[n,n]:=d[n+1]; od:
for n from 2 to M-1 do
for i from 1 to n-1 do T[n,i]:=T[n-1,i-1]+T[n-1,i]; od: od:
lprint("triangle:");
for n from 0 to M-1 do lprint(seq(T[n,i],i=0..n)); od:
lprint("row sums:");
lprint([seq( add(T[i,j],j=0..i), i=0..M-1)]);
end;
f(d);

t[n_, n_] := n*(n+1)/2; t[n_, 0] := n*(n+1)/2; t[n_, k_] := t[n, k] = t[n-1, k-1] + t[n-1, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 20 2014 *)

G.f. as triangle: (1+x-4*x*y+x*y^2+x^2*y^2)*y/((1-y)^2*(-x*y+1)^2*(-x*y-y+1)). - Robert Israel, Apr 04 2018

A232094 a(n) = A060130(A000217(n)); number of nonzero digits in factorial base representation (A007623) of 0+1+2+...+n.

Original entry on oeis.org

0, 1, 2, 1, 2, 3, 3, 2, 2, 4, 3, 2, 2, 3, 4, 1, 3, 5, 4, 4, 3, 5, 3, 3, 3, 4, 5, 2, 4, 4, 5, 3, 2, 5, 4, 3, 3, 4, 4, 3, 3, 5, 6, 5, 4, 5, 3, 3, 3, 4, 5, 3, 5, 6, 5, 3, 4, 6, 5, 4, 4, 5, 6, 3, 5, 6, 4, 4, 4, 5, 5, 4, 4, 5, 5, 4, 4, 4, 6, 5, 2, 6, 5, 3, 4, 4, 5

Offset: 0

Antti Karttunen, Nov 18 2013

The next 1 after a(1), a(3) and a(15) occurs at n=224, as A000217(224) = 25200 = 5 * 7!.

Antti Karttunen, Table of n, a(n) for n = 0..10000

Cf. A000217, A007623, A060130, A226061, A230410, A232095.

a[n_] := Module[{k = n*(n+1)/2, m = 2, r, s = {}}, While[{k, r} = QuotientRemainder[k, m]; k != 0|| r != 0, AppendTo[s, r]; m++]; Count[s, ?(# > 0 &)]]; Array[a, 100, 0] (* _Amiram Eldar, Feb 07 2024 *)

(define (A232094 n) (A060130 (A000217 n)))

a(n) = A060130(A000217(n)).

a(n) = A230410(A226061(n+1)). [Not a practical way to compute this sequence. Please see comments at A230410.]

A233450 Numbers n such that 3*T(n)+1 is a square, where T = A000217.

Original entry on oeis.org

0, 1, 6, 15, 64, 153, 638, 1519, 6320, 15041, 62566, 148895, 619344, 1473913, 6130878, 14590239, 60689440, 144428481, 600763526, 1429694575, 5946945824, 14152517273, 58868694718, 140095478159, 582740001360, 1386802264321, 5768531318886, 13727927165055

Offset: 1

Bruno Berselli, Dec 10 2013

For n>1, partial sums of A080872 starting from A080872(1).

			153 is in the sequence because 3*153*154/2+1 = 188^2.

Bruno Berselli, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (1,10,-10,-1,1).

Sequence A129444 gives n+1.
Cf. A000217, A080872, A129445 (square roots of 3*A000217(a(n))+1), A132596 (numbers m such that 3*A000217(m) is a square).
Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; m=0 for k=2; this sequence for k=3; A001652 for k=4; A129556 for k=5; A001921 for k=6.

LinearRecurrence[{1, 10, -10, -1, 1}, {0, 1, 6, 15, 64}, 30]

G.f.: x^2*(1 + 5*x - x^2 - x^3) / ((1 - x)*(1 - 10*x^2 + x^4)).
a(n) = a(n-1) +10*a(n-2) -10*a(n-3) -a(n-4) +a(n-5) for n>5, a(1)=0, a(2)=1, a(3)=6, a(4)=15, a(5)=64.
a(n) = -1/2 + ( (-3*(-1)^n + 2*sqrt(6))*(5 + 2*sqrt(6))^floor(n/2) - (3*(-1)^n + 2*sqrt(6))*(5 - 2*sqrt(6))^floor(n/2) )/12.

A242962 a(1) = a(2) = 0; for n >= 3: a(n) = (n*(n+1)/2) mod antisigma(n) = A000217(n) mod A024816(n).

Original entry on oeis.org

0, 0, 0, 1, 6, 3, 8, 15, 13, 18, 12, 28, 14, 24, 24, 31, 18, 39, 20, 42, 32, 36, 24, 60, 31, 42, 40, 56, 30, 72, 32, 63, 48, 54, 48, 91, 38, 60, 56, 90, 42, 96, 44, 84, 78, 72, 48, 124, 57, 93, 72, 98, 54, 120, 72, 120, 80, 90, 60, 168, 62, 96, 104, 127, 84

Offset: 1

Jaroslav Krizek, May 29 2014

nonn

A000217(n) = triangular numbers, A024816(n) = sum of numbers less than n which do not divide n.

a(n) = sigma(n) = A000203(n) for n = 5 and n>= 7 (see A242963).

			a(6) = 3 because A000217(6) mod A024816(6) = 21 mod 9 = 3.

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Cf. A000217, A002961, A024816, A242963, A243117, A243118.

[((n*(n+1)div 2) mod (n*(n+1)div 2-SumOfDivisors(n))): n in [3..1000]]

Array[If[# < 3, 0, Mod[PolygonalNumber@ #, Total@ Complement[Range@ #, Divisors@ #]]] &, 65] (* Michael De Vlieger, Jan 28 2020 *)

A252117 Irregular triangle read by row: T(n,k), n>=1, k>=1, in which column k lists the numbers of A000716 multiplied by A000330(k), and the first element of column k is in row A000217(k).

Original entry on oeis.org

1, 3, 9, 5, 22, 15, 51, 45, 108, 110, 14, 221, 255, 42, 429, 540, 126, 810, 1105, 308, 1479, 2145, 714, 30, 2640, 4050, 1512, 90, 4599, 7395, 3094, 270, 7868, 13200, 6006, 660, 13209, 22995, 11340, 1530, 21843, 39340, 20706, 3240, 55, 35581, 66045, 36960, 6630, 165, 57222, 109215, 64386, 12870, 495

Offset: 1

Omar E. Pol, Dec 14 2014

Gives an identity for sigma(n). Alternating sum of row n equals A000203(n), the sum of the divisors of n.
Row n has length A003056(n) hence column k starts in row A000217(k).
Column 1 is A000716, but here the offset is 1 not 0.
The 1st element of column k is A000330(k).
The 2nd element of column k is A059270(k).
The 3rd element of column k is A220443(k).
The partial sums of column k give the k-th column of A249120.
This triangle has been constructed after Mircea Merca's formula for A000203.
From Omar E. Pol, May 05 2022: (Start)
In the Honda-Yoda paper see "3. String theory and Riemann hypothesis". The coefficients that are mentioned in 3.11 are the first 16 terms of A000716, the coefficients that are mentioned in 3.12 are the first 5 terms of A000330, and the coefficients that are mentioned in 3.13 are the first 16 terms of A000203.
Another triangle with the same row lengths and whose alternating row sums give A000203 is A196020. (End)

			Triangle begins:
       1;
       3;
       9,      5;
      22,     15;
      51,     45;
     108,    110,     14;
     221,    255,     42;
     429,    540,    126;
     810,   1105,    308;
    1479,   2145,    714,     30;
    2640,   4050,   1512,     90;
    4599,   7395,   3094,    270;
    7868,  13200,   6006,    660;
   13209,  22995,  11340,   1530;
   21843,  39340,  20706,   3240,    55;
   35581,  66045,  36960,   6630,   165;
   57222, 109215,  64386,  12870,   495;
   90882, 177905, 110152,  24300,  1210;
  142769, 286110, 184926,  44370,  2805;
  221910, 454410, 305802,  79200,  5940;
  341649, 713845, 498134, 137970, 12155, 91;
...
For n = 6 the divisors of 6 are 1, 2, 3, 6, so the sum of the divisors of 6 is 1 + 2 + 3 + 6 = 12. On the other hand, the 6th row of the triangle is 108, 110, 14, so the alternating row sum is 108 - 110 + 14 = 12, equaling the sum of the divisors of 6.
For n = 15 the divisors of 15 are 1, 3, 5, 15, so the sum of the divisors of 15 is 1 + 3 + 5 + 15 = 24. On the other hand, the 15th row of the triangle is 21843, 39340, 20706, 3240, 55, so the alternating row sum is 21843 - 39340 + 20706 - 3240 + 55 = 24, equaling the sum of the divisors of 15.

Masazumi Honda and Takuya Yoda, String theory, N = 4 SYM and Riemann hypothesis, arXiv:2203.17091 [hep-th], (2022), pp. 5-6.
Index entries for sequences related to sigma(n)

Cf. A000203, A000217, A000330, A000716, A003056, A059270, A196020, A220443, A238442, A249120.

A003056(n) = (sqrtint(8*n+1)-1)\2;
A000330(n) = n*(n+1)*(2*n+1)/6;
A000217(n) = n*(n+1)/2;
A000716(n) = polcoef(1/eta('x+O('x^(n+1)))^3, n, x);
T(n, k) = A000330(k)*A000716(n-A000217(k));
row(n) = vector(A003056(n), k, T(n,k)); \\ Michel Marcus, Oct 29 2022

A000203(n) = Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k).

A274757 Numbers k such that 6*k+1 is a triangular number (A000217).

Original entry on oeis.org

0, 9, 15, 42, 54, 99, 117, 180, 204, 285, 315, 414, 450, 567, 609, 744, 792, 945, 999, 1170, 1230, 1419, 1485, 1692, 1764, 1989, 2067, 2310, 2394, 2655, 2745, 3024, 3120, 3417, 3519, 3834, 3942, 4275, 4389, 4740, 4860, 5229, 5355, 5742, 5874, 6279, 6417

Offset: 1

Colin Barker, Jul 04 2016

Numbers of the type floor(3*m*(m+1)/4) for which floor(3*m*(m+1)/4) = 3*floor(m*(m+1)/4). A014601 lists the values of m. - Bruno Berselli, Jan 13 2017

Numbers of the form 3*k*(4*k + 1) for k in Z. - Peter Bala, Nov 21 2024

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A000217, A014601.
Cf. A000096 (k+1), A074377 (2*k+1), A045943 (3*k+1), A274681 (4*k+1), A085787 (5*k+1).
Cf. similar sequences listed in A274830.

Table[3 (2 n - 1) (2 n + (-1)^n - 1)/4, {n, 1, 60}] (* Bruno Berselli, Jul 08 2016 *)
LinearRecurrence[{1,2,-2,-1,1},{0,9,15,42,54},50] (* Harvey P. Dale, Apr 13 2025 *)

PARI
```
isok(n) = ispolygonal(6*n+1, 3)
```

select(n->ispolygonal(6*n+1, 3), vector(7000, n, n-1))

concat(0, Vec(3*x^2*(3+2*x+3*x^2)/((1-x)^3*(1+x)^2) + O(x^60)))

G.f.: 3*x^2*(3 + 2*x + 3*x^2) / ((1 - x)^3*(1 + x)^2).
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n>5.
a(n) = 3*(2*n - 1)*(2*n + (-1)^n - 1)/4. Therefore:
a(n) = 3*n*(2*n - 1)/2 for n even,
a(n) = 3*(n-1)*(2*n - 1)/2 for n odd.

cp's OEIS Frontend

A137281 Numbers k such that T(k) is not divisible by T(i), 1 < i < k, where T(k) = k-th triangular number A000217(k).

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A138796 Least possible k > 0 with T(k) - T(j) = n, j > 0, where T(i) > 0 are the triangular numbers A000217.

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A181388 a(n) = Sum_{k=1..n} 2^T(k-1), where T = A000217 are the triangular numbers 0, 1, 3, 6, 10, ... . For n=0 we have the empty sum equal to 0.

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A215795 Numbers n such that 2^n-1 is a triangular number (A000217).

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A222403 Triangle read by rows: left and right edges are A000217, interior entries are filled in using the Pascal triangle rule.

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A232094 a(n) = A060130(A000217(n)); number of nonzero digits in factorial base representation (A007623) of 0+1+2+...+n.

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A233450 Numbers n such that 3*T(n)+1 is a square, where T = A000217.

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