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 1-5 of 5 results.

A136118 Least index m>0 such that A136117(n)-A000326(m) is again a pentagonal number.

Original entry on oeis.org

5, 4, 7, 12, 19, 17, 25, 20, 10, 28, 45, 42, 39, 17, 37, 21, 36, 35, 13, 33, 65, 28, 67, 32, 52, 40, 74, 31, 70, 85, 35, 16, 60, 70, 77, 68, 42, 30, 105, 76, 59, 26, 74, 49, 115, 19, 125, 115, 102, 110, 92, 56, 103, 29, 145, 100, 114, 77, 92, 47, 63, 108, 152, 95, 22, 116
Offset: 1

Views

Author

M. F. Hasler, Dec 25 2007

Keywords

Examples

			a(1)=5 is the least integer m>0 such that A136117(1)-P(m) is a pentagonal number, namely P(7)-P(5)=70-35=35=P(5).
a(2)=4 is the least integer m>0 such that A136117(2)-P(m) is a pentagonal number, namely P(8)-P(4)=92-22=70=P(7).
		

Crossrefs

Programs

  • PARI
    A136118vect(n,i=-1)=vector(n,k,until(0,for(j=2,#n=sum2sqr((i+=6)^2+1),n[j]%6==[5,5]||next;n=n[j];break(2)));n[1]\6+1) /* This uses sum2sqr(), cf. A133388. Below some simpler but much slower code. */
    my(P=A000326(n)=n*(3*n-1)/2,isPent(t)=P(sqrtint(t*2\3)+1)==t); for(i=1,299,for(j=1,(i+1)\sqrt(2),isPent(P(i)-P(j))&print1(j",")||next(2)))

A133215 Hexagonal numbers (A000384) which are sum of 2 other hexagonal numbers > 0.

Original entry on oeis.org

276, 703, 861, 1225, 2850, 3003, 4560, 5151, 8128, 10878, 11781, 12090, 12720, 13366, 14706, 15400, 16110, 18721, 21115, 22366, 24090, 24531, 26796, 29161, 29646, 31125, 32131, 33153, 36315, 38503, 39621, 40186, 42486, 45451, 47895
Offset: 1

Views

Author

Jonathan Vos Post, Dec 18 2007

Keywords

Comments

This is to A136117 as A000384 is to A000326. Duke and Schulze-Pillot (1990) proved that every sufficiently large integer (and hence every sufficiently large hexagonal number) can be written as the sum of three hexagonal numbers.

Examples

			hex(19) = 703 = 378 + 325 = hex(14) + hex(13).
hex(21) = 861 = 630 + 231 = hex(18) + hex(11).
hex(25) = 1225 = 1035 + 190 = hex(23) + hex(10).
hex(38) = 2850 = 2415 + 435 = hex(35) + hex(15).
hex(39) = 3003 = 2850 + 153 = hex(38) + hex(9) = 2415 + 435 + 153 = hex(35) + hex(15) + hex(9).
hex(48) = 4560 = 2415 + 2145 = hex(35) + hex(33).
		

Crossrefs

Programs

  • Mathematica
    With[{upto=60000},Select[Union[Total/@Subsets[Table[n(2n-1),{n, Ceiling[ (1+Sqrt[1+8upto])/4]}],{2}]],IntegerQ[(1+Sqrt[1+8#])/4]&&#<=upto&]] (* Harvey P. Dale, Jul 24 2011 *)

Formula

{x: x>0 and x in A000384 and x = A000384(i) + A000384(j) for i>0 and j>0}, where A000384 = {n*(2*n-1) for n > 0}.

Extensions

Added missing term 276 and a(8)-a(35) from Donovan Johnson, Sep 27 2008

A136346 Octagonal numbers which are the sums of exactly two positive octagonal numbers.

Original entry on oeis.org

560, 736, 1541, 3201, 5461, 6816, 7400, 9976, 11041, 11408, 13333, 14981, 15408, 15841, 19521, 21000, 21505, 25761, 28616, 30401, 41536, 45141, 50440, 51221, 52008, 54405, 56856, 61920, 63656, 65416, 69008, 75525, 76480, 81345, 82336, 85345, 87381, 89441
Offset: 1

Views

Author

Jonathan Vos Post, Dec 25 2007

Keywords

Comments

For sums of two positive octagonal numbers, see A136345. This is to octagonal numbers A000567 as A089982 is to triangular numbers A000217, as A009000 is to squares A000290, as A136117 is to pentagonal numbers A000326, as A133215 is to hexagonal numbers A000384, and as A117104 is to heptagonal numbers A000566. If Oc(a) + Oc(b) = Oc(c) then a(3a-2) + b(3b+2) = c(3c+2), so solving the quadratic equations for c we have (when an integer): c = (2 + sqrt(4 + 36a^2 + 36b^2 - 24a - 24b))/6.

Examples

			Where Oc(n) = A000567(n) = n-th octagonal number:
a(1) = 560 = Oc(14) = 280 + 280 = Oc(10) + Oc(10).
a(2) = 736 = Oc(16) = 560 + 176 = Oc(14) + Oc(8).
a(3) = 1541 = Oc(23) = 1408 + 133 = Oc(22) + Oc(7).
a(4) = 3201 = Oc(33) = 2465 + 736 = Oc(29) + Oc(16).
a(5) = 5461 = Oc(43) = 2821 + 2640 = Oc(31) + Oc(30).
		

Crossrefs

Programs

  • Mathematica
    Module[{nn=300,ono},ono=PolygonalNumber[8,Range[nn]];Union[Select[ Total/@ Tuples[ono,2],MemberQ[ono,#]&]]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Oct 26 2019 *)

Formula

A000567 INTERSECTION {A000567(i) + A000567(j), i, j > 0}. {i*(3*i-2)} INTERSECTION {i*(3*i-2) + j(3*j-2), i > 0}.

Extensions

Corrected and edited by B. D. Swan (bdswan(AT)gmail.com), Dec 20 2008

A136345 Sums of exactly two positive octagonal numbers A000567.

Original entry on oeis.org

2, 9, 16, 22, 29, 41, 42, 48, 61, 66, 73, 80, 86, 97, 104, 105, 117, 130, 134, 136, 141, 154, 161, 173, 177, 184, 192, 197, 198, 216, 226, 229, 233, 241, 246, 265, 266, 272, 281, 288, 290, 301, 309, 320, 321, 342, 345, 349, 352, 358, 362, 376, 381, 401, 406, 409
Offset: 1

Views

Author

Jonathan Vos Post, Dec 25 2007

Keywords

Comments

For octagonal numbers within this sequence, see: A136346. This is to octagonal numbers A000567 as A051533 is to triangular numbers A000217 and as A000404 is to squares A000290 and as A117104 is to heptagonal numbers A000566.

Crossrefs

Formula

{A000567(i) + A000567(j), i, j > 0}. {i*(3*i-2) + j(3*j-2), i, j > 0}.

Extensions

Missing term (80) from Giovanni Resta, Jun 19 2016

A133251 Heptagonal numbers A000566 which are the sum of two other heptagonal numbers > 0.

Original entry on oeis.org

697, 3186, 3744, 5221, 7209, 8323, 12496, 12852, 19228, 20566, 21022, 24850, 29539, 35224, 38254, 40768, 44023, 44689, 52345, 53802, 58293, 62173, 63760, 66178, 67815, 78057, 79834, 80730, 82537, 95746, 97713, 101707, 115240, 131905, 135373
Offset: 1

Views

Author

Jonathan Vos Post, Dec 19 2007

Keywords

Comments

This is to A000566 as A136117 is to A000326.
The sequence contains 12852 and 19751431167846, which are the smallest heptagonal numbers equal to twice another heptagonal number. - R. J. Mathar, Jan 13 2008

Examples

			Where hep(k) = k-th heptagonal number = A000566(k):
a(1) = 697 = hep(17) = 616 + 81 = hep(16) + hep(6).
a(2) = 3186 = hep(36) = 1782 + 1404 = hep(27) + hep(24).
a(3) = 3744 = hep(39) = 2673 + 1071 = hep(33) + hep(21).
a(4) = 5221 = hep(46) = 4347 + 874 = hep(42) + hep(19).
		

Crossrefs

Programs

  • Mathematica
    Module[{nn=1000,heps},heps=Table[(n(5n-3))/2,{n,nn}]; Select[ Union[ Total/@ Tuples[Take[heps,nn/2],2]],MemberQ[heps,#]&]] (* Harvey P. Dale, Dec 18 2015 *)

Formula

{x such that x in A000566 and x = A000566(i) + A000566(j) for i, j > 0 and where A000566(k) = k*(5*k-3)/2}.

Extensions

More terms from R. J. Mathar, Jan 13 2008
Showing 1-5 of 5 results.