A162623 -id:A162623 - cp's OEIS frontend

A162614 Triangle read by rows in which row n lists n+1 terms, starting with n, such that the difference between successive terms is equal to n^3 - 1.

0, 1, 1, 2, 9, 16, 3, 29, 55, 81, 4, 67, 130, 193, 256, 5, 129, 253, 377, 501, 625, 6, 221, 436, 651, 866, 1081, 1296, 7, 349, 691, 1033, 1375, 1717, 2059, 2401, 8, 519, 1030, 1541, 2052, 2563, 3074, 3585, 4096, 9, 737, 1465, 2193, 2921, 3649, 4377, 5105, 5833

Offset: 0

Omar E. Pol, Jul 15 2009

Note that the last term of the n-th row is the fourth power of n, A000583(n).

See also the triangles of A162615 and A162616.

			Triangle begins:
  0;
  1,   1;
  2,   9,  16;
  3,  29,  55,  81;
  4,  67, 130, 193, 256;
  5, 129, 253, 377, 501,  625;
  6, 221, 436, 651, 866, 1081, 1296;
  ...

Cf. A000583, A068601, A159797, A162609, A162610, A162611, A162612, A162613, A162615, A162616, A162622, A162623, A162624.

def A162614(n,k):
    return n+k*(n**3-1)
print([A162614(n,k) for n in range(20) for k in range(n+1)])
# R. J. Mathar, Oct 20 2009

Sum_{k=0..n} T(n,k) = n*(n^2-n+1)*(n+1)^2/2 (row sums). - R. J. Mathar, Jul 20 2009

T(n,k) = n + k*(n^3-1). - R. J. Mathar, Oct 20 2009

More terms from R. J. Mathar, Oct 20 2009

A162622 Triangle read by rows in which row n lists n+1 terms, starting with n, such that the difference between successive terms is equal to n^4 - 1.

Original entry on oeis.org

0, 1, 1, 2, 17, 32, 3, 83, 163, 243, 4, 259, 514, 769, 1024, 5, 629, 1253, 1877, 2501, 3125, 6, 1301, 2596, 3891, 5186, 6481, 7776, 7, 2407, 4807, 7207, 9607, 12007, 14407, 16807, 8, 4103, 8198, 12293, 16388, 20483, 24578, 28673, 32768, 9, 6569, 13129

Offset: 0

Omar E. Pol, Jul 15 2009

Note that the last term of the n-th row is the 5th power of n, A000584(n).

See also the triangles of A162623 and A162624.

			Triangle begins:
  0;
  1,    1;
  2,   17,    32;
  3,   83,   163,   243;
  4,  259,   514,   769,  1024;
  5,  629,  1253,  1877,  2501,  3125;
  6, 1301,  2596,  3891,  5186,  6481,  7776;
  7, 2407,  4807,  7207,  9607, 12007, 14407, 16807;
  8, 4103,  8198, 12293, 16388, 20483, 24578, 28673, 32768;
  9, 6569, 13129, 19689, 26249, 32809, 39369, 45929, 52489, 59049; etc.

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Cf. A000583, A000584, A123865, A159797, A162609, A162610, A162611, A162612, A162613, A162614, A162615, A162616, A162623, A162624.

/* Triangle: */ [[n+k*(n^4-1): k in [0..n]]: n in [0..10]]; // Bruno Berselli, Dec 14 2012

A162622 := proc(n,k) n+k*(n^4-1) ; end proc: seq(seq( A162622(n,k),k=0..n),n=0..15) ; # R. J. Mathar, Feb 11 2010

Flatten[Table[NestList[#+n^4-1&,n,n],{n,0,9}]] (* Harvey P. Dale, Jun 23 2013 *)

Sum_{k=0..n} T(n,k) = n*(n+1)*(1+n^4)/2 (row sums). [R. J. Mathar, Jul 20 2009]

7th and later rows from R. J. Mathar, Feb 11 2010

A177342 a(n) = (4n^3-3n^2+5*n-3)/3.

Original entry on oeis.org

1, 9, 31, 75, 149, 261, 419, 631, 905, 1249, 1671, 2179, 2781, 3485, 4299, 5231, 6289, 7481, 8815, 10299, 11941, 13749, 15731, 17895, 20249, 22801, 25559, 28531, 31725, 35149, 38811, 42719, 46881, 51305, 55999, 60971, 66229, 71781, 77635

Offset: 1

Bruno Berselli, May 06 2010 - Nov 27 2010

This sequence is related to the fourth powers (A000583) by n^4 = n*a(n) - Sum_{i=1..n-1} a(i) - (n-1), with n>1.

Also, n*a(n) - Sum_{i=1..n-1} a(i) provides the first column of A162624 and the second column of A162622 (or A162623). - Bruno Berselli, revised Dec 14 2012

B. Berselli, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

First differences: 2*A084849.
Partial sums: A178073.
Cf. A174723, A162622-A162624.

[(4*n^3-3*n^2+5*n-3)/3: n in [1..39]]; // Bruno Berselli, Aug 24 2011

I:=[1,9,31,75]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Aug 19 2013

CoefficientList[Series[(1 + 5 x + x^2 + x^3) / (1 - x)^4, {x, 0, 50}], x] (* Vincenzo Librandi, Aug 19 2013 *)
Table[(4 n^3 - 3 n^2 + 5 n - 3)/3, {n, 1, 40}] (* Bruno Berselli, Feb 17 2015 *)
LinearRecurrence[{4,-6,4,-1},{1,9,31,75},40] (* Harvey P. Dale, Jul 31 2021 *)

a(n)=(4*n^3-3*n^2+5*n-3)/3 \\ Charles R Greathouse IV, Jun 23 2011

G.f.: x*(1 + 5*x + x^2 + x^3)/(1 - x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
a(n) - a(-n) = 2*A004006(2n).
a(n) + a(-n) = -A002522(n).
a(n) = 1 + (n-1)*(4*n^2+n+6)/3 = 2*A174723(n)-1.

Formulae added and revised by Bruno Berselli, Feb 17 2015

A162624 Triangle read by rows in which row n lists n terms, starting with n^4 + n - 1, such that the difference between successive terms is equal to n^4 - 1 = A123865(n).

Original entry on oeis.org

1, 17, 32, 83, 163, 243, 259, 514, 769, 1024, 629, 1253, 1877, 2501, 3125, 1301, 2596, 3891, 5186, 6481, 7776, 2407, 4807, 7207, 9607, 12007, 14407, 16807, 4103, 8198, 12293, 16388, 20483, 24578, 28673, 32768, 6569, 13129, 19689, 26249, 32809

Offset: 1

Omar E. Pol, Jul 12 2009

Note that the last term of the n-th row is the 5th power of n, A000584(n).

See also the triangles of A162622 and A162623.

			Triangle begins:
     1;
    17,   32;
    83,  163,  243;
   259,  514,  769, 1024;
   629, 1253, 1877, 2501, 3125;
  1301, 2596, 3891, 5186, 6481, 7776;
  ...

Nathaniel Johnston, Table of n, a(n) for n = 1..10000

Cf. A000584, A123865, A159797, A162609, A162610, A162611, A162612, A162613, A162614, A162615, A162616, A162622, A162623.

A162624 := proc(n,k) return n+k*(n^4-1): end: seq(seq(A162624(n,k), k=1..n), n=1..10); # Nathaniel Johnston, Apr 30 2011

Table[NestList[#+n^4-1&,n^4+n-1,n-1],{n,10}]//Flatten (* Harvey P. Dale, Apr 28 2022 *)

Row sums: n*(n^5 + n^4 + n - 1)/2. - R. J. Mathar, Jul 20 2009

cp's OEIS Frontend

A162614 Triangle read by rows in which row n lists n+1 terms, starting with n, such that the difference between successive terms is equal to n^3 - 1.

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A162622 Triangle read by rows in which row n lists n+1 terms, starting with n, such that the difference between successive terms is equal to n^4 - 1.

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A177342 a(n) = (4*n^3-3*n^2+5*n-3)/3.

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A162624 Triangle read by rows in which row n lists n terms, starting with n^4 + n - 1, such that the difference between successive terms is equal to n^4 - 1 = A123865(n).

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Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A177342 a(n) = (4n^3-3n^2+5*n-3)/3.