A101165 -id:A101165 - cp's OEIS frontend

A005449 Second pentagonal numbers: a(n) = n(3n + 1)/2.

0, 2, 7, 15, 26, 40, 57, 77, 100, 126, 155, 187, 222, 260, 301, 345, 392, 442, 495, 551, 610, 672, 737, 805, 876, 950, 1027, 1107, 1190, 1276, 1365, 1457, 1552, 1650, 1751, 1855, 1962, 2072, 2185, 2301, 2420, 2542, 2667, 2795, 2926, 3060, 3197, 3337, 3480

Offset: 0

N. J. A. Sloane

Number of edges in the join of the complete graph and the cycle graph, both of order n, K_n * C_n. - Roberto E. Martinez II, Jan 07 2002
Also number of cards to build an n-tier house of cards. - Martin Wohlgemuth, Aug 11 2002
The modular form Delta(q) = q*Product_{n>=1} (1-q^n)^24 = q*(1 + Sum_{n>=1} (-1)^n*(q^(n*(3*n-1)/2)+q^(n*(3*n+1)/2)))^24 = q*(1 + Sum_{n>=1} A033999(n)*(q^A000326(n)+q^a(n)))^24. - Jonathan Vos Post, Mar 15 2006
Row sums of triangle A134403.
Bisection of A001318. - Omar E. Pol, Aug 22 2011
Sequence found by reading the line from 0 in the direction 0, 7, ... and the line from 2 in the direction 2, 15, ... in the square spiral whose vertices are the generalized pentagonal numbers, A001318. - Omar E. Pol, Sep 08 2011
A general formula for the n-th second k-gonal number is given by T(n, k) = n*((k-2)*n+k-4)/2, n>=0, k>=5. - Omar E. Pol, Aug 04 2012
Partial sums give A006002. - Denis Borris, Jan 07 2013
A002260 is the following array A read by antidiagonals:
  0,  1,  2,  3,  4,  5, ...
  0,  1,  2,  3,  4,  5, ...
  0,  1,  2,  3,  4,  5, ...
  0,  1,  2,  3,  4,  5, ...
  0,  1,  2,  3,  4,  5, ...
  0,  1,  2,  3,  4,  5, ...
and a(n) is the hook sum: Sum_{k=0..n} A(n,k) + Sum_{r=0..n-1} A(r,n). - R. J. Mathar, Jun 30 2013
From Klaus Purath, May 13 2021: (Start)
This sequence and A000326 provide all integers m such that 24*m + 1 is a square. The union of the two sequences is A001318.
If A is a sequence satisfying the recurrence t(n) = 3*t(n-1) - 2*t(n-2) with the initial values either A(0) = 1, A(1) = n + 2 or A(0) = -1, A(1) = n - 1, then a(n) = (A(i)^2 - A(i-1)*A(i+1))/2^i + n^2 for i>0. (End)
a(n+1) is the number of Dyck paths of size (3,3n+2), i.e., the number of NE lattice paths from (0,0) to (3,3n+2) which stay above the line connecting these points. - Harry Richman, Jul 13 2021
Binomial transform of [0, 2, 3, 0, 0, 0, ...], being a(n) = 2*binomial(n,1) + 3*binomial(n,2). a(3) = 15 = [0, 2, 3, 0] dot [1, 3, 3, 1] = [0 + 6 + 9 + 0]. - Gary W. Adamson, Dec 17 2022
a(n) is the sum of longest side length of all nondegenerate integer-sided triangles with shortest side length n and middle side length (n + 1), n > 0. - Torlach Rush, Feb 04 2024

			From _Omar E. Pol_, Aug 22 2011: (Start)
Illustration of initial terms:
                                               O
                                             O O
                                 O         O O O
                               O O       O O O O
                     O       O O O     O O O O O
                   O O     O O O O     O O O O O
           O     O O O     O O O O     O O O O O
         O O     O O O     O O O O     O O O O O
    O    O O     O O O     O O O O     O O O O O
    O    O O     O O O     O O O O     O O O O O
    -    ---     -----     -------     ---------
    2     7        15         26           40
(End)

Henri Cohen, A Course in Computational Algebraic Number Theory, vol. 138 of Graduate Texts in Mathematics, Springer-Verlag, 2000.

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
A. O. L. Atkin and F. Morain, Elliptic Curves and Primality Proving, Math. Comp., Vol. 61, No. 203 (1993), pp. 29-68.
Charles H. Conley and Valentin Ovsienko, Quiddities of polygon dissections and the Conway-Coxeter frieze equation, arXiv:2107.01234 [math.CO], 2021.
Leonhard Euler, De mirabilibus proprietatibus numerorum pentagonalium, Acta Academiae Scientiarum Imperialis Petropolitanae, Vol. 1780: I, pp. 56-75.
Leonhard Euler, Observatio de summis divisorum, Novi Commentarii academiae scientiarum Petropolitanae, Vol. 5, pp. 59-74.
Leonhard Euler, An observation on the sums of divisors, arXiv:math/0411587 [math.HO], 2004-2009, p. 8.
Leonhard Euler, On the remarkable properties of the pentagonal numbers, arXiv:math/0505373 [math.HO], 2005.
Alfred Hoehn, Illustration of initial terms of A000326, A005449, A045943, A115067.
D. Suprijanto and I. W. Suwarno, Observation on Sums of Powers of Integers Divisible by 3k-1, Applied Mathematical Sciences, Vol. 8, No. 45 (2014), pp. 2211-2217.
Martin Wohlgemuth, Pentagon, Kartenhaus und Summenzerlegung.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A016789 (first differences), A006002 (partial sums).
Cf. A000217, A000320, A000326, A001318, A033568, A049451, A101165, A101166.
The generalized pentagonal numbers b*n+3*n*(n-1)/2, for b = 1 through 12, form sequences A000326, this sequence, A045943, A115067, A140090, A140091, A059845, A140672-A140675, A151542.
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488 (this sequence is the case k=3).
Cf. numbers of the form n*((2*k+1)*n+1)/2 listed in A022289 (this sequence is the case k=1).

[n*(3*n + 1) / 2: n in [0..40]]; // Vincenzo Librandi, May 02 2011

A005449:=n->n*(3*n + 1)/2; seq(A005449(k), k=0..100); # Wesley Ivan Hurt, Oct 11 2013

Table[n (3 n + 1)/2, {n, 0, 100}] (* Zak Seidov, Jan 31 2012 *)

{a(n) = n * (3*n + 1) / 2} /* Michael Somos, Jul 18 2003 */

a(n) = A110449(n, 1) for n>0.
G.f.: x*(2+x)/(1-x)^3. E.g.f.: exp(x)*(2*x + 3*x^2/2). a(n) = n*(3*n + 1)/2. a(-n) = A000326(n). - Michael Somos, Jul 18 2003
a(n) = A001844(n) - A000217(n+1) = A101164(n+2,2) for n>0. - Reinhard Zumkeller, Dec 03 2004
a(n) = Sum_{j=1..n} n+j. - Zerinvary Lajos, Sep 12 2006
a(n) = A126890(n,n). - Reinhard Zumkeller, Dec 30 2006
a(n) = 2*C(3*n,4)/C(3*n,2), n>=1. - Zerinvary Lajos, Jan 02 2007
a(n) = A000217(n) + A000290(n). - Zak Seidov, Apr 06 2008
a(n) = a(n-1) + 3*n - 1 for n>0, a(0)=0. - Vincenzo Librandi, Nov 18 2010
a(n) = A129267(n+5,n). - Philippe Deléham, Dec 21 2011
a(n) = 2*A000217(n) + A000217(n-1). - Philippe Deléham, Mar 25 2013
a(n) = A130518(3*n+1). - Philippe Deléham, Mar 26 2013
a(n) = (12/(n+2)!)*Sum_{j=0..n} (-1)^(n-j)*binomial(n,j)*j^(n+2). - Vladimir Kruchinin, Jun 04 2013
a(n) = floor(n/(1-exp(-2/(3*n)))) for n>0. - Richard R. Forberg, Jun 22 2013
a(n) = Sum_{i=1..n} (3*i - 1) for n >= 1. - Wesley Ivan Hurt, Oct 11 2013 [Corrected by Rémi Guillaume, Oct 24 2024]
a(n) = (A000292(6*n+k+1)-A000292(k))/(6*n+1) - A000217(3*n+k+1), for any k >= 0. - Manfred Arens, Apr 26 2015
Sum_{n>=1} 1/a(n) = 6 - Pi/sqrt(3) - 3*log(3) = 0.89036376976145307522... . - Vaclav Kotesovec, Apr 27 2016
a(n) = A000217(2*n) - A000217(n). - Bruno Berselli, Sep 21 2016
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*Pi/sqrt(3) + 4*log(2) - 6. - Amiram Eldar, Jan 18 2021
From Klaus Purath, May 13 2021: (Start)
Partial sums of A016789 for n > 0.
a(n) = 3*n^2 - A000326(n).
a(n) = A000326(n) + n.
a(n) = A002378(n) + A000217(n-1) for n >= 1. [Corrected by Rémi Guillaume, Aug 14 2024] (End)
From Klaus Purath, Jul 14 2021: (Start)
b^2 = 24*a(n) + 1 we get by b^2 = (a(n+1) - a(n-1))^2 = (a(2*n)/n)^2.
a(2*n) = n*(a(n+1) - a(n-1)), n > 0.
a(2*n+1) = n*(a(n+1) - a(n)). (End)
A generalization of Lajos' formula, dated Sep 12 2006, follows. Let SP(k,n) = the n-th second k-gonal number. Then SP(2k+1,n) = Sum_{j=1..n} (k-1)*n+j+k-2. - Charlie Marion, Jul 13 2024
a(n) = Sum_{k = 0..3*n} (-1)^(n+k+1) * binomial(k, 2) * binomial(3*n+k, 2*k). - Peter Bala, Nov 03 2024
For integer m, (6*m + 1)^2*a(n) + a(m) = a((6*m+1)*n + m). - Peter Bala, Jan 09 2025

A185787 Sum of first k numbers in column k of the natural number array A000027; by antidiagonals.

Original entry on oeis.org

1, 7, 25, 62, 125, 221, 357, 540, 777, 1075, 1441, 1882, 2405, 3017, 3725, 4536, 5457, 6495, 7657, 8950, 10381, 11957, 13685, 15572, 17625, 19851, 22257, 24850, 27637, 30625, 33821, 37232, 40865, 44727, 48825, 53166, 57757, 62605, 67717, 73100, 78761, 84707, 90945, 97482, 104325, 111481, 118957, 126760, 134897, 143375

Offset: 1

Clark Kimberling, Feb 03 2011

This is one of many interesting sequences and arrays that stem from the natural number array A000027, of which a northwest corner is as follows:
  1....2.....4.....7...11...16...22...29...
  3....5.....8....12...17...23...30...38...
  6....9....13....18...24...31...39...48...
  10...14...19....25...32...40...49...59...
  15...20...26....33...41...50...60...71...
  21...27...34....42...51...61...72...84...
  28...35...43....52...62...73...85...98...
Blocking out all terms below the main diagonal leaves columns whose sums comprise A185787.  Deleting the main diagonal and then summing give A185787.  Analogous treatments to the left of the main diagonal give A100182 and A101165.  Further sequences obtained directly from this array are easily obtained using the following formula for the array: T(n,k)=n+(n+k-2)(n+k-1)/2.
Examples:
row 1:  A000124
row 2:  A022856
row 3:  A016028
row 4:  A145018
row 5:  A077169
col 1:  A000217
col 2:  A000096
col 3:  A034856
col 4:  A055998
col 5:  A046691
col 6:  A052905
col 7:  A055999
diag. (1,5,...) ...... A001844
diag. (2,8,...) ...... A001105
diag. (4,12,...)...... A046092
diag. (7,17,...)...... A056220
diag. (11,23,...) .... A132209
diag. (16,30,...) .... A054000
diag. (22,38,...) .... A090288
diag. (3,9,...) ...... A058331
diag. (6,14,...) ..... A051890
diag. (10,20,...) .... A005893
diag. (15,27,...) .... A097080
diag. (21,35,...) .... A093328
antidiagonal sums:  (1,5,15,34,...)=A006003=partial sums of A002817.
Let S(n,k) denote the n-th partial sum of column k.  Then
S(n,k)=n*(n^2+3k*n+3*k^2-6*k+5)/6.
S(n,1)=n(n+1)(n+2)/6
S(n,2)=n(n+1)(n+5)/6
S(n,3)=n(n+2)(n+7)/6
S(n,4)=n(n^2+12n+29)/6
S(n,5)=n(n+5)(n+10)/6
S(n,6)=n(n+7)(n+11)/6
S(n,7)=n(n+10)(n+11)/6
Weight array of T: A144112
Accumulation array of T: A185506
Second rectangular sum array of T: A185507
Third rectangular sum array of T: A185508
Fourth rectangular sum array of T: A185509

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000027, A185788, A100182, A101165, A079824.

[n*(7*n^2-6*n+5)/6: n in [1..50]]; // Vincenzo Librandi, Jul 04 2012

f[n_,k_]:=n+(n+k-2)(n+k-1)/2;
s[k_]:=Sum[f[n,k],{n,1,k}];
Factor[s[k]]
Table[s[k],{k,1,70}]  (* A185787 *)
CoefficientList[Series[(3*x^2+3*x+1)/(1-x)^4,{x,0,50}],x] (* Vincenzo Librandi, Jul 04 2012 *)

a(n)=n*(7*n^2-6*n+5)/6.

G.f.: x*(3*x^2+3*x+1)/(1-x)^4. - Vincenzo Librandi, Jul 04 2012

Edited by Clark Kimberling, Feb 25 2023

A101164 Triangle read by rows: Delannoy numbers minus binomial coefficients.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 3, 7, 3, 0, 0, 4, 15, 15, 4, 0, 0, 5, 26, 43, 26, 5, 0, 0, 6, 40, 94, 94, 40, 6, 0, 0, 7, 57, 175, 251, 175, 57, 7, 0, 0, 8, 77, 293, 555, 555, 293, 77, 8, 0, 0, 9, 100, 455, 1079, 1431, 1079, 455, 100, 9, 0, 0, 10, 126, 668, 1911, 3191, 3191, 1911, 668, 126, 10, 0

Offset: 0

Reinhard Zumkeller, Dec 03 2004

			Triangle begins as:
  0
  0, 0;
  0, 1,  0;
  0, 2,  2,   0;
  0, 3,  7,   3,   0;
  0, 4, 15,  15,   4,   0;
  0, 5, 26,  43,  26,   5,  0;
  0, 6, 40,  94,  94,  40,  6, 0;
  0, 7, 57, 175, 251, 175, 57, 7, 0;

Reinhard Zumkeller, Rows n = 0..100 of table, flattened
Eric Weisstein's World of Mathematics, Delannoy Number
Eric Weisstein's World of Mathematics, Binomial Coefficient
Index entries for triangles and arrays related to Pascal's triangle

Cf. A001477, A005449, A008288, A094706, A101165, A101166, A225413.

a101164 n k = a101164_tabl !! n !! k
a101164_row n = a101164_tabl !! n
a101164_tabl = zipWith (zipWith (-)) a008288_tabl a007318_tabl
-- Reinhard Zumkeller, Jul 30 2013

A101164:= func< n,k | (&+[Binomial(n-k,j)*Binomial(k,j)*2^j: j in [0..n-k]]) - Binomial(n,k) >;
[A101164(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 17 2021

T[n_, k_]:= Hypergeometric2F1[-k, k-n, 1, 2] - Binomial[n, k];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 17 2021 *)

def T(n,k): return simplify(hypergeometric([-n+k, -k], [1], 2)) - binomial(n,k)
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Sep 17 2021

T(n, k) = A008288(n, k) - binomial(n, k), 0<=k<=n, where binomial=A007318.
T(n,2) = A005449(n-2) for n>1;
T(n,3) = A101165(n-3) for n>2;
T(n,4) = A101166(n-4) for n>3;
Sum_{k=0..n} T(n, k) = A094706(n).
From G. C. Greubel, Sep 17 2021: (Start)
T(n, k) = Sum_{j=0..n-k} binomial(n-k, j)*binomial(k, j)*2^j - binomial(n,k).
T(n, 1) = n-1, n > 0. (End)

A101166 a(n) = (15n^4 + 22n^3 + 45n^2 + 14n) / 24.

Original entry on oeis.org

0, 4, 26, 94, 251, 555, 1079, 1911, 3154, 4926, 7360, 10604, 14821, 20189, 26901, 35165, 45204, 57256, 71574, 88426, 108095, 130879, 157091, 187059, 221126, 259650, 303004, 351576, 405769, 466001, 532705, 606329, 687336, 776204, 873426

Offset: 0

Reinhard Zumkeller, Dec 03 2004

a(n) = A001846(n) - A000332(n+4) = A101164(n+4,4) for n > 0.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5, -10, 10, -5, 1).

Cf. A005449, A101165.

[(15*n^4+22*n^3+45*n^2+14*n) / 24: n in [0..40]]; // Vincenzo Librandi, Dec 26 2010

Table[(15n^4+22n^3+45n^2+14n)/24,{n,0,40}] (* or *) LinearRecurrence[ {5,-10,10,-5,1},{0,4,26,94,251},40] (* Harvey P. Dale, Oct 12 2012 *)

a(n)=(15*n^4+22*n^3+45*n^2+14*n)/24 \\ Charles R Greathouse IV, Oct 16 2015

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), with a(0)=0, a(1)=4, a(2)=26, a(3)=94, a(4)=251. - Harvey P. Dale, Oct 12 2012

G.f.: (-x^4-4*x^3-6*x^2-4*x)/(x-1)^5. - Harvey P. Dale, Oct 12 2012

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A005449 Second pentagonal numbers: a(n) = n*(3*n + 1)/2.

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A185787 Sum of first k numbers in column k of the natural number array A000027; by antidiagonals.

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A101164 Triangle read by rows: Delannoy numbers minus binomial coefficients.

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A101166 a(n) = (15*n^4 + 22*n^3 + 45*n^2 + 14*n) / 24.

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A005449 Second pentagonal numbers: a(n) = n(3n + 1)/2.

A101166 a(n) = (15n^4 + 22n^3 + 45n^2 + 14n) / 24.