A006414 -id:A006414 - cp's OEIS frontend

A107980 Triangle read by rows: T(n,k) = (n+2)(k+1)(k+2)(2n-k+2)(2n-k+3)/24.

1, 5, 9, 14, 30, 40, 30, 70, 105, 125, 55, 135, 216, 280, 315, 91, 231, 385, 525, 630, 686, 140, 364, 624, 880, 1100, 1260, 1344, 204, 540, 945, 1365, 1755, 2079, 2310, 2430, 285, 765, 1360, 2000, 2625, 3185, 3640, 3960, 4125, 385, 1045, 1881, 2805, 3740, 4620, 5390, 6006, 6435, 6655

Offset: 0

Emeric Deutsch, Jun 12 2005

Kekulé numbers for certain benzenoids.

			Triangle begins:
    1;
    5,    9;
   14,   30,   40;
   30,   70,  105,  125;
   55,  135,  216,  280,  315;
   91,  231,  385,  525,  630,  686;
  140,  364,  624,  880, 1100, 1260, 1344;
  204,  540,  945, 1365, 1755, 2079, 2310, 2430;
  285,  765, 1360, 2000, 2625, 3185, 3640, 3960, 4125;
  385, 1045, 1881, 2805, 3740, 4620, 5390, 6006, 6435, 6655;
  506, 1386, 2520, 3800, 5130, 6426, 7616, 8640, 9450, 10010, 10296;

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 237, K{B(n,3,l)}).

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000330, A000389, A006414, A006858.

T:=proc(n,k) if k<=n then 1/24*(n+2)*(k+1)*(k+2)*(2*n-k+2)*(2*n-k+3) else 0 fi end: for n from 0 to 9 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

T[n_, k_]:= (1/6)*(n+2)*Binomial[k+2, 2]*Binomial[2*n-k+3, 2];
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 14 2021 *)

def A107980(n,k): return (n+2)*(k+1)*(k+2)*(2*n-k+2)*(2*n-k+3)/24
flatten([[A107980(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Dec 14 2021

T(n, k) = (n+2)*(k+1)*(k+2)*(2*n-k+2)*(2*n-k+3)/24.
T(n, 0) = A000330(n+1).
T(n, n) = A006414(n).
Sum_{k=0..n} T(n, k) = A006858(n+1).
T(n, n-1) = 5*binomial(n+4, 5) = 5*A000389(n+4). - G. C. Greubel, Dec 14 2021

A354968 Triangle read by rows: T(n, k) = nk(n+k)*(n-k)/6.

Original entry on oeis.org

1, 4, 5, 10, 16, 14, 20, 35, 40, 30, 35, 64, 81, 80, 55, 56, 105, 140, 154, 140, 91, 84, 160, 220, 256, 260, 224, 140, 120, 231, 324, 390, 420, 405, 336, 204, 165, 320, 455, 560, 625, 640, 595, 480, 285, 220, 429, 616, 770, 880, 935, 924, 836, 660, 385, 286, 560, 810, 1024

Offset: 2

Ali Sada and Yifan Xie, Jun 14 2022

Given a Pythagorean triple (a,b,c), define S = c^4 - a^4 - b^4. Using Euclid's parameterization (a = 2*n*k, b = n^2 - k^2, c = n^2 + k^2), substituting to get S in terms of n and k gives S = 8*n^2*k^2*((n^2 - k^2))^2, which is a multiple of 288; T(n, k) = sqrt(S/288) = n*k*(n^2 - k^2)/6 = n*k*(n+k)*(n-k)/6.

			Triangle begins:
  n/k   1    2    3    4    5    6    7
  2     1;
  3     4,   5;
  4    10,  16,  14;
  5    20,  35,  40,  30;
  6    35,  64,  81,  80,  55;
  7    56, 105, 140, 154, 140,  91;
  8    84, 160, 220, 256, 260, 224, 140;
  ...
For n = 3, k = 2, a = 5, b = 12, c = 13. T(3, 2) = sqrt((13^4 - 5^4 - 12^4)/288) = 5.

James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Page 72.

M. F. Hasler, Table of n, a(n) for n = 2..1000, May 08 2025
Wikipedia, Pythagorean Triple.
Index entries related to Pythagorean Triples.

Cf. A120070 (b leg), A055096 (c hypotenuse).

Cf. A006414 (row sums), A000292 (column 1), A077414 (column 2), A000330 (diagonal), A107984 (transpose), A210440 (diagonal which begins with 4).

T[n_,k_]:=n*k(n^2-k^2)/6; Table[T[n,k],{n,2,11},{k,n-1}]//Flatten (* Stefano Spezia, Jul 11 2025 *)

apply( {A354968(n, k=0)=k|| k=n-1-(1-n=ceil(sqrt(8*n-7)/2+.5))*(2-n)\2; k*(n-k)*n*(n+k)\6}, [2..66]) \\ M. F. Hasler, May 08 2025

G.f.: x^2*y*(1 + x*y - 4*x^2*y + x^3*y + x^4*y^2)/((1 - x)^4*(1 - x*y)^4). - Stefano Spezia, Jul 11 2025

A195166 Numbers expressible as 2^a - 2^b, with 0 <= b < a, such that n^a - n^b is divisible by 2^a - 2^b for all n.

Original entry on oeis.org

1, 2, 6, 12, 30, 24, 60, 120, 252, 240, 504, 16380, 32760, 65520

Offset: 1

Michel Marcus, Dec 21 2012

= 2^1  - 2^0. (n^1  - n^0)/1     : A000027
= 2^2  - 2^1. (n^2  - n^1)/2     : A000217
= 2^3  - 2^1. (n^3  - n^1)/6     : A000292
= 2^4  - 2^2. (n^4  - n^2)/12    : A002415
= 2^5  - 2^1. (n^5  - n^1)/30    : A033455
= 2^5  - 2^3. (n^5  - n^3)/24    : A006414
= 2^6  - 2^2. (n^6  - n^2)/60    : A213547
= 2^7  - 2^3. (n^7  - n^3)/120   : A114239
= 2^8  - 2^2. (n^8  - n^2)/252   :
= 2^8  - 2^4. (n^8  - n^4)/240   : A078876
= 2^9  - 2^3. (n^9  - n^3)/504   :
= 2^14 - 2^2. (n^14 - n^2)/16380 :
= 2^15 - 2^3. (n^15 - n^3)/32760 :
= 2^16 - 2^4. (n^16 - n^4)/65520 :
Comment from Tomohiro Yamada, Oct 05 2022: (Start)
"The Mod Set Stanford University" and Carl Pomerance independently noted that the completeness of this sequence follows from a result of Schinzel on primitive prime factors of sequences a^n - b^n in the remark to a problem of Harry Ruderman asking whether if 2^a - 2^b divides 3^a - 3^b, then 2^a - 2^b belongs to this sequence.
The Mod Set verified that if a > b, 2^a - 2^b divides 3^a - 3^b, but 2^a - 2^b does not belong to this sequence, then a - b > 1900. Ram Murty and Kumar Murty proved that there are only finitely many natural numbers a, b such that 2^a - 2^b divides 3^a - 3^b.
Qi Sun and Ming Zhi Zhang also showed that if a > b and n^a - n^b is divisible by 2^a - 2^b for all n, then 2^a - 2^b belongs to this sequence. (End)

			a(3) = 6 belongs to this sequence since (n^3 - n)/6 = C(n+1, 3) = A000292(n-1).

M. Ram Murty and V. Kumar Murty, On a Problem of Ruderman, Amer. Math. Monthly 118 (2011), 644-650, available from the first author's website.
Harry Ruderman, Problem E2468, Amer. Math. Monthly 81 (1974), p. 405.
A. Schinzel, On primitive prime factors of a^n - b^n, Proc. Cambridge Phil. Soc. 58 (1962), 556-562.
Qi Sun and Ming Zhi Zhang, Pairs where 2^a-2^b divides n^a-n^b for all n, Proc. Amer. Math. Soc. 93 (1985), 218-220.
The Mod Set Stanford University and Carl Pomerance, When 2^m - 2^n divides 3^m - 3^n, remarks to Problem E2468*, Amer. Math. Monthly 84 (1977), 59-60.
W. Y. Velez, When 2^m - 2^n divides 3^m - 3^n, remarks to Problem E2468, Amer. Math. Monthly 83 (1976), 288-289.

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A107980 Triangle read by rows: T(n,k) = (n+2)(k+1)(k+2)(2n-k+2)(2n-k+3)/24.

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A354968 Triangle read by rows: T(n, k) = nk(n+k)*(n-k)/6.

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A195166 Numbers expressible as 2^a - 2^b, with 0 <= b < a, such that n^a - n^b is divisible by 2^a - 2^b for all n.

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cp's OEIS Frontend

A107980 Triangle read by rows: T(n,k) = (n+2)*(k+1)*(k+2)*(2*n-k+2)*(2*n-k+3)/24.

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Programs

Maple

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Formula

A354968 Triangle read by rows: T(n, k) = n*k*(n+k)*(n-k)/6.

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Mathematica

PARI

Formula

A195166 Numbers expressible as 2^a - 2^b, with 0 <= b < a, such that n^a - n^b is divisible by 2^a - 2^b for all n.

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A107980 Triangle read by rows: T(n,k) = (n+2)(k+1)(k+2)(2n-k+2)(2n-k+3)/24.

A354968 Triangle read by rows: T(n, k) = nk(n+k)*(n-k)/6.