A264443 -id:A264443 - cp's OEIS frontend

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

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

Peter Bala, Nov 13 2015

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.

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

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

[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 );
```

Table[n (n + 7) (n + 14)/6, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)

vector(100, n, n--; n*(n+7)*(n+14)/6) \\ Altug Alkan, Nov 15 2015

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

Peter Bala, Nov 13 2015

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.

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

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

[n*(n+11)*(n+22)/6: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015

seq( n*(n + 11)*(n + 22)/6, n = 0..40 );

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 *)

vector(100, n, n--;  n*(n+11)*(n+22)/6) \\ Altug Alkan, Nov 15 2015

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

Peter Bala, Nov 13 2015

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.

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

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

[n*(n+5)*(n+10)*(n+15)/24: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015

seq( n*(n + 5)*(n + 10)*(n + 15)/24, n = 0..40 );

Table[n (n + 5) (n + 10) (n + 15)/24, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)

vector(100, n, n--;  n*(n+5)*(n+10)*(n+15)/24) \\ Altug Alkan, Nov 15 2015

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

A264447 a(n) = n(n + 7)(n + 14)*(n + 21)/24.

Original entry on oeis.org

0, 110, 276, 510, 825, 1235, 1755, 2401, 3190, 4140, 5270, 6600, 8151, 9945, 12005, 14355, 17020, 20026, 23400, 27170, 31365, 36015, 41151, 46805, 53010, 59800, 67210, 75276, 84035, 93525, 103785, 114855, 126776, 139590, 153340, 168070, 183825, 200651, 218595, 237705, 258030

Offset: 0

Peter Bala, Nov 13 2015

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 = 7.

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A007310, A264443, A264444, A264445, A264446, A264448, A264449, A264450.

[n*(n+7)*(n+14)*(n+21)/24: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015

seq( n*(n + 7)*(n + 14)*(n + 21)/24, n = 0..40 );

Table[n (n + 7) (n + 14) (n + 21)/24, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)
Table[Times@@(n+7*Range[0,3])/24,{n,0,40}] (* or *) LinearRecurrence[{5,-10,10,-5,1},{0,110,276,510,825},50] (* Harvey P. Dale, May 01 2017 *)

vector(100, n, n--; n*(n+7)*(n+14)*(n+21)/24) \\ Altug Alkan, Nov 15 2015

O.g.f.: x*(110 - 274*x + 230*x^2 - 65*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

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

Peter Bala, Nov 13 2015

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.

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

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

[n*(n+11)*(n+22)*(n+33)/24: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015

seq( n*(n + 11)*(n + 22)*(n + 33)/24, n = 0..38 );

Table[n (n + 11) (n + 22) (n + 33)/24, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)

vector(100, n, n--; n*(n+11)*(n+22)*(n+33)/24) \\ Altug Alkan, Nov 15 2015

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

Peter Bala, Nov 13 2015

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.

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

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

[n*(n+7)*(n+14)*(n+21)*(n+28)/120: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015

seq( n*(n + 7)*(n + 14)*(n + 21)*(n + 28)/120, n = 0..35 );

Table[n (n + 7) (n + 14) (n + 21) (n + 28)/120, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)

vector(100, n, n--; n*(n+7)*(n+14)*(n+21)*(n+28)/120) \\ Altug Alkan, Nov 15 2015

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

Peter Bala, Nov 13 2015

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.

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

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

[n*(n+11)*(n+22)*(n+33)*(n+44)/120: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015

seq( n*(n + 11)*(n + 22)*(n + 33)*(n + 44)/120, n = 0..33 );

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 *)

vector(100, n, n--; n*(n+11)*(n+22)*(n+33)*(n+44)/120) \\ Altug Alkan, Nov 15 2015

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

cp's OEIS Frontend

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

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

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Magma

Maple

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

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Author

Keywords

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Programs

Magma

Maple

Mathematica

PARI

Formula

A264447 a(n) = n*(n + 7)*(n + 14)*(n + 21)/24.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

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

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

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

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

A264447 a(n) = n(n + 7)(n + 14)*(n + 21)/24.

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

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

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