cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A264443 a(n) = n*(n + 5)*(n + 10)/6.

Original entry on oeis.org

0, 11, 28, 52, 84, 125, 176, 238, 312, 399, 500, 616, 748, 897, 1064, 1250, 1456, 1683, 1932, 2204, 2500, 2821, 3168, 3542, 3944, 4375, 4836, 5328, 5852, 6409, 7000, 7626, 8288, 8987, 9724, 10500, 11316, 12173, 13072, 14014, 15000
Offset: 0

Views

Author

Peter Bala, Nov 13 2015

Keywords

Comments

It is well-known, and easy to prove, that the product of 3 consecutive integers n*(n + 1)*(n + 2) is divisible by 6. It can be shown that the product of 3 integers in arithmetic progression n*(n + r)*(n + 2*r) is divisible by 6 if and only if r is not divisible by 2 or 3 (see A007310 for these numbers). This is the case r = 5.

Links

Crossrefs

Cf. A007310, A264444, A264445, A264446, A264447, A264448, A264449, A264450.

Programs

  • Magma
    [n*(n+5)*(n+10)/6: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015
  • Maple
    seq( n*(n + 5)*(n + 10)/6, n = 1..40 );
  • Mathematica
    Table[n (n + 5) (n + 10)/6, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)
  • PARI
    vector(100, n, n--; n*(n+5)*(n+10)/6) \\ Altug Alkan, Nov 15 2015

Formula

O.g.f.: x*(6*x^2 - 16*x + 11)/(1 - x)^4.
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>3. - Vincenzo Librandi, Nov 16 2015

A264444 a(n) = n*(n + 7)*(n + 14)/6.

Original entry on oeis.org

0, 20, 48, 85, 132, 190, 260, 343, 440, 552, 680, 825, 988, 1170, 1372, 1595, 1840, 2108, 2400, 2717, 3060, 3430, 3828, 4255, 4712, 5200, 5720, 6273, 6860, 7482, 8140, 8835, 9568, 10340, 11152, 12005, 12900, 13838, 14820, 15847, 16920
Offset: 0

Views

Author

Peter Bala, Nov 13 2015

Keywords

Comments

It is well-known, and easy to prove, that the product of 3 consecutive integers n*(n + 1)*(n + 2) is divisible by 6. It can be shown that the product of 3 integers in arithmetic progression n*(n + r)*(n + 2*r) is divisible by 6 if and only if r is not divisible by 2 or 3 (see A007310 for these numbers). This is the case r = 7.

Links

Crossrefs

Cf. A007310, A264443, A264445, A264446, A264447, A264448, A264449, A264450.

Programs

  • Magma
    [n*(n+7)*(n+14)/6: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015
  • Maple
    seq( n*(n + 7)*(n + 14)/6, n = 0..40 );
  • Mathematica
    Table[n (n + 7) (n + 14)/6, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)
  • PARI
    vector(100, n, n--; n*(n+7)*(n+14)/6) \\ Altug Alkan, Nov 15 2015

Formula

O.g.f.: x*(13*x^2 - 32*x + 20)/(1 - x)^4.
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>3. - Vincenzo Librandi, Nov 16 2015

A264445 a(n) = n*(n + 11)*(n + 22)/6.

Original entry on oeis.org

0, 46, 104, 175, 260, 360, 476, 609, 760, 930, 1120, 1331, 1564, 1820, 2100, 2405, 2736, 3094, 3480, 3895, 4340, 4816, 5324, 5865, 6440, 7050, 7696, 8379, 9100, 9860, 10660, 11501, 12384, 13310, 14280, 15295, 16356, 17464, 18620, 19825, 21080
Offset: 0

Views

Author

Peter Bala, Nov 13 2015

Keywords

Comments

It is well-known, and easy to prove, that the product of 3 consecutive integers n*(n + 1)*(n + 2) is divisible by 6. It can be shown that the product of 3 integers in arithmetic progression n*(n + r)*(n + 2*r) is divisible by 6 if and only if r is not divisible by 2 or 3 (see A007310 for these numbers). This is the case r = 11.

Links

Crossrefs

Cf. A007310, A264443, A264444, A264446, A264447, A264448, A264449, A264450.

Programs

  • Magma
    [n*(n+11)*(n+22)/6: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015
  • Maple
    seq( n*(n + 11)*(n + 22)/6, n = 0..40 );
  • Mathematica
    Table[n (n + 11) (n + 22)/6, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)
LinearRecurrence[{4,-6,4,-1},{0,46,104,175},50] (* Harvey P. Dale, Dec 11 2018 *)
  • PARI
    vector(100, n, n--;  n*(n+11)*(n+22)/6) \\ Altug Alkan, Nov 15 2015

Formula

O.g.f.: x*(35*x^2 - 80*x + 46)/(1 - x)^4.
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>3. - Vincenzo Librandi, Nov 16 2015

A264446 a(n) = n*(n + 5)*(n + 10)*(n + 15)/24.

Original entry on oeis.org

0, 44, 119, 234, 399, 625, 924, 1309, 1794, 2394, 3125, 4004, 5049, 6279, 7714, 9375, 11284, 13464, 15939, 18734, 21875, 25389, 29304, 33649, 38454, 43750, 49569, 55944, 62909, 70499, 78750, 87699, 97384, 107844, 119119, 131250, 144279, 158249, 173204, 189189, 206250
Offset: 0

Views

Author

Peter Bala, Nov 13 2015

Keywords

Comments

It is well-known, and easy to prove, that the product of 4 consecutive integers n*(n + 1)*(n + 2)*(n + 3) is divisible by 4!. It can be shown that the product of 4 integers in arithmetic progression n*(n + r)*(n + 2*r)*(n + 3*r) is divisible by 4! if and only if r is not divisible by 2 or 3 (see A007310 for these numbers). This is the case r = 5.

Links

Crossrefs

Cf. A007310, A264443, A264444, A264445, A264447, A264448, A264449, A264450.

Programs

  • Magma
    [n*(n+5)*(n+10)*(n+15)/24: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015
  • Maple
    seq( n*(n + 5)*(n + 10)*(n + 15)/24, n = 0..40 );
  • Mathematica
    Table[n (n + 5) (n + 10) (n + 15)/24, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)
  • PARI
    vector(100, n, n--;  n*(n+5)*(n+10)*(n+15)/24) \\ Altug Alkan, Nov 15 2015

Formula

O.g.f.: x*(4 - 3*x)*(7*x^2 - 17*x + 11)/(1 - x)^5.
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5) for n>4. - Vincenzo Librandi, Nov 16 2015

A264448 a(n) = n*(n + 11)*(n + 22)*(n + 33)/24.

Original entry on oeis.org

0, 391, 910, 1575, 2405, 3420, 4641, 6090, 7790, 9765, 12040, 14641, 17595, 20930, 24675, 28860, 33516, 38675, 44370, 50635, 57505, 65016, 73205, 82110, 91770, 102225, 113516, 125685, 138775, 152830, 167895, 184016, 201240, 219615, 239190, 260015, 282141, 305620, 330505
Offset: 0

Views

Author

Peter Bala, Nov 13 2015

Keywords

Comments

It is well-known, and easy to prove, that the product of 4 consecutive integers n*(n + 1)*(n + 2)*(n + 3) is divisible by 4!. It can be shown that the product of 4 integers in arithmetic progression n*(n + r)*(n + 2*r)*(n + 3*r) is divisible by 4! if and only if r is not divisible by 2 or 3 (see A007310 for these numbers). This is the case r = 11.

Links

Crossrefs

Cf. A007310, A264443, A264444, A264445, A264446, A264447, A264449, A264450.

Programs

  • Magma
    [n*(n+11)*(n+22)*(n+33)/24: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015
  • Maple
    seq( n*(n + 11)*(n + 22)*(n + 33)/24, n = 0..38 );
  • Mathematica
    Table[n (n + 11) (n + 22) (n + 33)/24, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)
  • PARI
    vector(100, n, n--; n*(n+11)*(n+22)*(n+33)/24) \\ Altug Alkan, Nov 15 2015

Formula

O.g.f.: x*(391 - 1045*x + 935*x^2 - 280*x^3)/(1 - x)^5.
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5) for n>4. - Vincenzo Librandi, Nov 16 2015

A264449 a(n) = n*(n + 7)*(n + 14)*(n + 21)*(n + 28)/120.

Original entry on oeis.org

0, 638, 1656, 3162, 5280, 8151, 11934, 16807, 22968, 30636, 40052, 51480, 65208, 81549, 100842, 123453, 149776, 180234, 215280, 255398, 301104, 352947, 411510, 477411, 551304, 633880, 725868, 828036, 941192, 1066185, 1203906, 1355289, 1521312, 1702998, 1901416, 2117682
Offset: 0

Views

Author

Peter Bala, Nov 13 2015

Keywords

Comments

It is well-known, and easy to prove, that the product of 5 consecutive integers n*(n + 1)*(n + 2)*(n + 3)*(n + 4) is divisible by 5!. It can be shown that the product of 5 integers in arithmetic progression n*(n + r)*(n + 2*r)*(n + 3*r)*(n + 4*r) is divisible by 5! if and only if r is not divisible by 2, 3 or 5 (see A007775 for these numbers). This is the case r = 7.

Links

Crossrefs

Cf. A007775, A264443, A264444, A264445, A264446, A264447, A264448, A264450.

Programs

  • Magma
    [n*(n+7)*(n+14)*(n+21)*(n+28)/120: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015
  • Maple
    seq( n*(n + 7)*(n + 14)*(n + 21)*(n + 28)/120, n = 0..35 );
  • Mathematica
    Table[n (n + 7) (n + 14) (n + 21) (n + 28)/120, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)
  • PARI
    vector(100, n, n--; n*(n+7)*(n+14)*(n+21)*(n+28)/120) \\ Altug Alkan, Nov 15 2015

Formula

O.g.f.: x*(351*x^4 - 1612*x^3 + 2796*x^2 - 2172*x + 638)/(1 - x)^6.
a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6), for n>5. - Vincenzo Librandi, Nov 16 2015

A264450 a(n) = n*(n + 11)*(n + 22)*(n + 33)*(n + 44)/120.

Original entry on oeis.org

0, 3519, 8372, 14805, 23088, 33516, 46410, 62118, 81016, 103509, 130032, 161051, 197064, 238602, 286230, 340548, 402192, 471835, 550188, 638001, 736064, 845208, 966306, 1100274, 1248072, 1410705, 1589224, 1784727, 1998360, 2231318, 2484846, 2760240, 3058848, 3382071
Offset: 0

Views

Author

Peter Bala, Nov 13 2015

Keywords

Comments

It is well-known, and easy to prove, that the product of 5 consecutive integers n*(n + 1)*(n + 2)*(n + 3)*(n + 4) is divisible by 5!. It can be shown that the product of 5 integers in arithmetic progression n*(n + r)*(n + 2*r)*(n + 3*r)*(n + 4*r) is divisible by 5! if and only if r is not divisible by 2, 3 or 5 (see A007775 for these numbers). This is the case r = 11.

Links

Crossrefs

Cf. A007775, A264443, A264444, A264445, A264446, A264447, A264448, A264449.

Programs

  • Magma
    [n*(n+11)*(n+22)*(n+33)*(n+44)/120: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015
  • Maple
    seq( n*(n + 11)*(n + 22)*(n + 33)*(n + 44)/120, n = 0..33 );
  • Mathematica
    Table[n (n + 11) (n + 22) (n + 33) (n + 44)/120, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)
LinearRecurrence[{6,-15,20,-15,6,-1},{0,3519,8372,14805,23088,33516},40] (* Harvey P. Dale, Nov 27 2015 *)
  • PARI
    vector(100, n, n--; n*(n+11)*(n+22)*(n+33)*(n+44)/120) \\ Altug Alkan, Nov 15 2015

Formula

O.g.f.: x*(2408*x^4 - 10542*x^3 + 17358*x^2 - 12742*x + 3519)/(1 - x)^6.
a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6) for n>5. - Vincenzo Librandi, Nov 16 2015
Showing 1-7 of 7 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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