A123296 -id:A123296 - cp's OEIS frontend

A262221 a(n) = 25n(n + 1)/2 + 1.

1, 26, 76, 151, 251, 376, 526, 701, 901, 1126, 1376, 1651, 1951, 2276, 2626, 3001, 3401, 3826, 4276, 4751, 5251, 5776, 6326, 6901, 7501, 8126, 8776, 9451, 10151, 10876, 11626, 12401, 13201, 14026, 14876, 15751, 16651, 17576, 18526, 19501, 20501, 21526, 22576, 23651

Offset: 0

Bruno Berselli, Sep 15 2015

Also centered 25-gonal (or icosipentagonal) numbers.
This is the case k=25 of the formula (k*n*(n+1) - (-1)^k + 1)/2. See table in Links section for similar sequences.
For k=2*n, the formula shown above gives A011379.
Primes in sequence: 151, 251, 701, 1951, 3001, 4751, 10151, 12401, ...

E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 51 (23rd row of the table).

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Table of numbers of the form (k*n*(n+1)-(-1)^k+1)/2.
Eric Weisstein's World of Mathematics, Centered Polygonal Numbers.
Index entries for sequences related to centered polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A008607 (first differences), A011379, A034856, A054254, A080956, A123296, A186349, A255184.
Cf. centered polygonal numbers listed in A069190.
Similar sequences of the form (k*n*(n+1) - (-1)^k + 1)/2 with -1 <= k <= 26: A000004, A000124, A002378, A005448, A005891, A028896, A033996, A035008, A046092, A049598, A060544, A064200, A069099, A069125, A069126, A069128, A069130, A069132, A069174, A069178, A080956, A124080, A163756, A163758, A163761, A164136, A173307.

Magma
```
[25*n*(n+1)/2+1: n in [0..50]];
```

Table[25 n (n + 1)/2 + 1, {n, 0, 50}]
25*Accumulate[Range[0,50]]+1 (* or *) LinearRecurrence[{3,-3,1},{1,26,76},50] (* Harvey P. Dale, Jan 29 2023 *)

PARI
```
vector(50, n, n--; 25*n*(n+1)/2+1)
```
Sage
```
[25*n*(n+1)/2+1 for n in (0..50)]
```

G.f.: (1 + 23*x + x^2)/(1 - x)^3.
a(n) = a(-n-1) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = A123296(n) + 1.
a(n) = A000217(5*n+2) - 2.
a(n) = A034856(5*n+1).
a(n) = A186349(10*n+1).
a(n) = A054254(5*n+2) with n>0, a(0)=1.
a(n) = A000217(n+1) + 23*A000217(n) + A000217(n-1) with A000217(-1)=0.
Sum_{i>=0} 1/a(i) = 1.078209111... = 2*Pi*tan(Pi*sqrt(17)/10)/(5*sqrt(17)).
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=0} a(n)/n! = 77*e/2.
Sum_{n>=0} (-1)^(n+1) * a(n)/n! = 23/(2*e). (End)
E.g.f.: exp(x)*(2 + 50*x + 25*x^2)/2. - Elmo R. Oliveira, Dec 24 2024

A061793 a(n) = 25n(n + 1)/2 + 3.

Original entry on oeis.org

3, 28, 78, 153, 253, 378, 528, 703, 903, 1128, 1378, 1653, 1953, 2278, 2628, 3003, 3403, 3828, 4278, 4753, 5253, 5778, 6328, 6903, 7503, 8128, 8778, 9453, 10153, 10878, 11628, 12403, 13203, 14028, 14878, 15753, 16653, 17578, 18528, 19503, 20503, 21528, 22578, 23653

Offset: 0

Jason Earls, Jun 22 2001

"If m is a triangular number, then so are 9*m+1, 25*m+3 and 49*m+6. (Euler, 1775)", see Burton in References. Note that A060544 is the same as 9*m+1 when m is triangular and that 9*(m*(m+1)/2)+1 is another formula for it.
9*m+1, 25*m+3 and 49*m+6 are special cases of the identity A000290(2*r + 1)*A000217(s) + A000217(r) = A000217((2*r + 1)*s + r). - Bruno Berselli, Mar 01 2018
Complementing the previous comment, with T(n) = A000217(n), 4*T(s)+1+s = T(2*s+1), 16*T(s)+3+2s = T(4*s+2) and 36*T(s)+6+3s = T(6*s+3) are special cases of the identity A000290(2*r)*T(s) + T(r) + r*s = T(2*r*s + r). - Charlie Marion, Mar 28 2018

D. M. Burton, Elementary Number Theory, Allyn and Bacon, Inc. Boston, MA, 1976, p. 17.

Harry J. Smith, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A000290, A060544, A123296.

List([0..40],n->25*(n*(n+1)/2)+3); # Muniru A Asiru, Mar 30 2018

[seq(25*(n*(n+1)/2)+3,n=0..40)]; # Muniru A Asiru, Mar 30 2018

25*Accumulate[Range[0,40]]+3 (* Harvey P. Dale, Aug 26 2013 *)

PARI
```
a(n) = 25*n*(n + 1)/2 + 3
```

a(n) = 25*A000217(n) + 3 = A123296(n) + 3.
From Elmo R. Oliveira, Oct 23 2024: (Start)
G.f.: (3 + 19*x + 3*x^2)/(1 - x)^3.
E.g.f.: (3 + 25*x + 25*x^2/2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. (End)

Corrected by T. D. Noe, Oct 25 2006

A162940 a(n) = binomial(n+1,2)*6^2.

Original entry on oeis.org

0, 36, 108, 216, 360, 540, 756, 1008, 1296, 1620, 1980, 2376, 2808, 3276, 3780, 4320, 4896, 5508, 6156, 6840, 7560, 8316, 9108, 9936, 10800, 11700, 12636, 13608, 14616, 15660, 16740, 17856, 19008, 20196, 21420, 22680, 23976, 25308, 26676, 28080, 29520, 30996

Offset: 0

Zerinvary Lajos, Jul 18 2009, Jul 19 2009

Number of n permutations (n>=2) of 7 objects s, t, u, v, z, x, y with repetition allowed, containing n-2 u's. Example: If n=2 then n-2 = zero (0) u, a(1)=36 because we have ss, st, sv, sz, sx, sy, ts, tt, tv, tz, tx, ty, vs, vt, vv, vz, vx, vy, zs, zt, zv, zz, zx, zy, xs, xt, xv, xz, xx, xy, ys, yt, yv, yz, yx, yy. If n=3 then n-2 = one (1) u, a(2) = 108, >> ssu, stu, etc. If n=4 then n-2 = two (2) u, a(2)= 216, >> ssuu, stuu, ..., txuu, etc. If n=5 then n-2 = three (3) u, a(3)=360, >> ssuuu, stuuu, ..., txuuu, etc.

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A002378, A046092, A027468, A035008, A123296.

Cf. A049598, A163758.

Table[Binomial[n + 1, 2]*6^2, {n, 0, 58}]

a(n)=18*n*(n+1) \\ Charles R Greathouse IV, Jun 16 2017

From Amiram Eldar, Sep 01 2022: (Start)
Sum_{n>=1} 1/a(n) = 1/18.
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2)/9 - 1/18. (End)
From Amiram Eldar, Feb 22 2023: (Start)
a(n) = 18*n*(n+1) = 36*A000217(n) = 18*A002378(n).
Product_{n>=1} (1 - 1/a(n)) = -(18/Pi)*cos(sqrt(11)*Pi/6).
Product_{n>=1} (1 + 1/a(n)) = (18/Pi)*cos(sqrt(7)*Pi/6). (End)
From Elmo R. Oliveira, Nov 29 2024: (Start)
G.f.: 36*x/(1-x)^3.
E.g.f.: 18*x*(2 + x)*exp(x).
a(n) = 3*A049598(n) = 2*A163758(n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A162942 a(n) = binomial(n+1,2)*7^2.

Original entry on oeis.org

0, 49, 147, 294, 490, 735, 1029, 1372, 1764, 2205, 2695, 3234, 3822, 4459, 5145, 5880, 6664, 7497, 8379, 9310, 10290, 11319, 12397, 13524, 14700, 15925, 17199, 18522, 19894, 21315, 22785, 24304, 25872, 27489, 29155, 30870, 32634, 34447, 36309

Offset: 0

Zerinvary Lajos, Jul 18 2009

Number of n permutations (n>=2) of 8 objects r, s, t, u, v, z, x, y with repetition allowed, containing n-2 u's.

			If n=2 then n-2=zero (0) u, a(1) = 49 because we have sr, tr, vr, zr, xr, yr, rs, rt, rv, rz, rx, ry, ss, st, sv, sz, sx, sy, ts, tt, tv, tz, tx, ty, vs, vt, vv, vz, vx, vy, zs, zt, zv, zz, zx, zy, xs, xt, xv, xz, xx, xy, ys, yt, yv, yz, yx, yy. If n=3 then n-2 = one (1) u, a(2) = 147 >> ssu, stu, etc.. Tf n=4 then n-2 = two (2) u, a(2) = 294 >> ssuu, stuu, ..., txuu, etc.. If n=5 then n-2 = three (3) u, a(3) = 490 >> rsuuu, stuuu, ..., rxuuu, etc..

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A027468, A027469, A035008, A046092, A123296, A162940.

Table[Binomial[n + 1, 2]*7^2, {n, 0, 58}]

a(n)=49*binomial(n+1,2) \\ Charles R Greathouse IV, May 02 2014

a(n) = A027469(n+2). - R. J. Mathar, Jul 18 2009
G.f.: -49*x/(x-1)^3. - R. J. Mathar, May 02 2014
From Amiram Eldar, Sep 04 2022: (Start)
Sum_{n>=1} 1/a(n) = 2/49.
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*(2*log(2)-1)/49. (End)
From Elmo R. Oliveira, Dec 27 2024: (Start)
E.g.f.: 49*exp(x)*x*(2 + x)/2.
a(n) = 49*A000217(n) = 49*n*(n+1)/2.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A305838 Triangle read by rows: T(0,0)= 1; T(n,k)= T(n-1,k) + 5*T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.

Original entry on oeis.org

1, 1, 1, 5, 1, 10, 1, 15, 25, 1, 20, 75, 1, 25, 150, 125, 1, 30, 250, 500, 1, 35, 375, 1250, 625, 1, 40, 525, 2500, 3125, 1, 45, 700, 4375, 9375, 3125, 1, 50, 900, 7000, 21875, 18750, 1, 55, 1125, 10500, 43750, 65625, 15625, 1, 60, 1375, 15000, 78750, 175000, 109375

Offset: 0

Shara Lalo, Jun 11 2018

The numbers in rows of the triangle are along skew diagonals pointing top-right in center-justified triangle given in A013612 ((1+5*x)^n). The coefficients in the expansion of 1/(1-x-5x^2) are given by the sequence generated by the row sums.

If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 2.7913327..., when n approaches infinity. The row sums are A015440 (generalized Fibonacci numbers).

			Triangle begins:
  1;
  1;
  1,  5;
  1, 10;
  1, 15,   25;
  1, 20,   75;
  1, 25,  150,   125;
  1, 30,  250,   500;
  1, 35,  375,  1250,    625;
  1, 40,  525,  2500,   3125;
  1, 45,  700,  4375,   9375,    3125;
  1, 50,  900,  7000,  21875,   18750;
  1, 55, 1125, 10500,  43750,   65625,   15625;
  1, 60, 1375, 15000,  78750,  175000,  109375;
  1, 65, 1650, 20625, 131250,  393750,  437500,   78125;
  1, 70, 1950, 27500, 206250,  787500, 1312500,  625000;
  1, 75, 2275, 35750, 309375, 1443750, 3281250, 2812500,  390625;
  1, 80, 2625, 45500, 446875, 2475000, 7218750, 9375000, 3515625;

Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 72, 380, 381.

Shara Lalo, Skew diagonals in triangle A013612
Shara Lalo, Right-justified triangle

Row sums give A015440.
Cf. A000012 (column 0), A008587 (column 1), A123296 (column 2), A141480 (column 3).
Cf. A013612.

t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0,  t[n - 1, k] + 5 t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 12}, {k, 0, Floor[n/2]}] // Flatten

G.f.: 1/(1 - t*x - 5*t^2).

cp's OEIS Frontend

A262221 a(n) = 25*n*(n + 1)/2 + 1.

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A061793 a(n) = 25*n*(n + 1)/2 + 3.

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A162940 a(n) = binomial(n+1,2)*6^2.

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A162942 a(n) = binomial(n+1,2)*7^2.

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A305838 Triangle read by rows: T(0,0)= 1; T(n,k)= T(n-1,k) + 5*T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.

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A262221 a(n) = 25n(n + 1)/2 + 1.

A061793 a(n) = 25n(n + 1)/2 + 3.