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 1-6 of 6 results.

A222082 Self-convolution square of A090845.

Original entry on oeis.org

1, 2, 5, 10, 20, 40, 67, 126, 203, 354, 571, 908, 1486, 2250, 3586, 5322, 8186, 12234, 17976, 26970, 38435, 57024, 80805, 116376, 165914, 232352, 332196, 456154, 645469, 885826, 1225998, 1692686, 2290512, 3168986, 4242896, 5805526, 7782803, 10459912, 14110205
Offset: 0

Views

Author

Paul D. Hanna, Feb 06 2013

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 5*x^2 + 10*x^3 + 20*x^4 + 40*x^5 + 67*x^6 +...
Let G(x) = A(x)^(1/2) denote the g.f. of A090845:
G(x) = 1 + x + 2*x^2 + 3*x^3 + 5*x^4 + 9*x^5 + 10*x^6 + 20*x^7 + 22*x^8 + 40*x^9 + 51*x^10 + 67*x^11 + 114*x^12 + 126*x^13 + 203*x^14 +...
then the coefficients of G(x)^2 and G(x)^3 begin:
G(x)^2: [1, 2, 5, 10, 20, 40, 67, 126, 203, 354, 571, 908, 1486, ...];
G(x)^3: [1, 3, 9, 22, 51, 114, 230, 468, 885, 1674, 3045, 5418, ..];
where the sorted union of these coefficients yield sequence A090845.

Links

Crossrefs

Cf. A090845, A222083.

Programs

  • PARI
    {a(n)=local(A=[1, 1]); for(i=1, #binary(3*n+1), A=vecsort(concat(Vec(Ser(A)^2), Vec(Ser(A)^3)))); Vec(Ser(A)^2)[n+1]}
for(n=0, 60, print1(a(n), ", "))

A222083 Self-convolution cube of A090845.

Original entry on oeis.org

1, 3, 9, 22, 51, 114, 230, 468, 885, 1674, 3045, 5418, 9560, 16341, 27912, 46383, 76794, 125205, 201580, 322980, 508710, 800495, 1241190, 1916682, 2935492, 4456617, 6747393, 10101532, 15105042, 22378362, 33035166, 48520809, 70776711, 103072393, 148899756
Offset: 0

Views

Author

Paul D. Hanna, Feb 06 2013

Keywords

Comments

A090846 gives the positions of where the terms of this sequence are found in A090845.

Examples

			G.f.: A(x) = 1 + 3*x + 9*x^2 + 22*x^3 + 51*x^4 + 114*x^5 + 230*x^6 +...
Let G(x) = A(x)^(1/3)  denote the g.f. of A090845:
G(x) = 1 + x + 2*x^2 + 3*x^3 + 5*x^4 + 9*x^5 + 10*x^6 + 20*x^7 + 22*x^8 + 40*x^9 + 51*x^10 + 67*x^11 + 114*x^12 + 126*x^13 + 203*x^14 +...
then the coefficients of G(x)^2 and G(x)^3 begin:
G(x)^2: [1, 2, 5, 10, 20, 40, 67, 126, 203, 354, 571, 908, 1486, ...];
G(x)^3: [1, 3, 9, 22, 51, 114, 230, 468, 885, 1674, 3045, 5418, ..];
where the sorted union of these coefficients yield sequence A090845.

Links

Crossrefs

Cf. A090845, A090846, A222082.

Programs

  • PARI
    {a(n)=local(A=[1, 1]); for(i=1, #binary(3*n+1), A=vecsort(concat(Vec(Ser(A)^2), Vec(Ser(A)^3)))); Vec(Ser(A)^3)[n+1]}
for(n=0, 60, print1(a(n), ", "))

A090846 Positions of the terms of A090845^3 in A090845, where A090845 is equal to the union of the self-convolutions A090845^2 and A090845^3, in ascending order by size, starting with A090845(0)=1.

Original entry on oeis.org

1, 3, 5, 8, 10, 12, 15, 17, 19, 22, 24, 27, 29, 31, 34, 36, 38, 41, 43, 45, 48, 50, 53, 55, 57, 60, 62, 64, 67, 69, 72, 74, 76, 79, 81, 83, 86, 88, 90, 93, 95, 98, 100, 102, 105, 107, 109, 112, 114, 117, 119, 121, 124, 126, 128, 131, 133, 135, 138, 140, 143, 145, 147, 150
Offset: 0

Views

Author

Paul D. Hanna, Dec 09 2003

Keywords

Comments

What is the value of limit a(n)/n ? Example: a(12000)/12000 = 2.3758333...

Examples

			a(4)=10 since A090845^3(4)=A090845(10)=51, where
A090845={1,1,2,3,5,9,10,20,22,40,51,...} and
A090845^3={1,3,9,22,51,114,230,468,885,1674,3045,5418,...}.

Links

Crossrefs

Cf. A090845, A222082, A222083.

Formula

A090845(a(n)) = A222083(n) for n>=0, where A222083 is the self-convolution cube of A090845.

A090848 Positions of the terms of A090847^4 in A090847, where A090847 is equal to the union of the self-convolutions A090847^2 and A090847^4 when ordered by size.

Original entry on oeis.org

1, 3, 6, 8, 11, 13, 16, 19, 21, 24, 26, 29, 32, 34, 37, 40, 42, 45, 47, 50, 53, 55, 58, 60, 63, 66, 68, 71, 73, 76, 79, 81, 84, 86, 89, 92, 94, 97, 99, 102, 105, 107, 110, 112, 115, 118, 120, 123, 126, 128, 131, 133, 136, 139, 141, 144, 146, 149, 152, 154, 157, 159, 162
Offset: 0

Views

Author

Paul D. Hanna, Dec 09 2003

Keywords

Comments

Given A090847(m)=A090847^4(n), then what is the limit m/n as n grows? Example: at n=2000, m/n=3202/2000=2.616, at n=3000, m/n=7849/3000=2.6163...

Examples

			a(4)=11 since A090847^4(4)=A090847(11)=117, where
A090847={1,1,2,4,5,12,14,22,44,50,88,117,...} and
A090847^4={1,4,14,44,117,308,740,1700,3822,...}.

Crossrefs

Cf. A090845, A090847.

A262990 G.f. A(x) satisfies: a([n/r^2]) = [x^n] A(x)^2/x and a([n/r^3]) = [x^n] A(x)^3/x^2, for n>=1, where r^2 + r^3 = 1.

Original entry on oeis.org

1, 1, 2, 3, 5, 9, 10, 20, 22, 40, 51, 67, 114, 126, 203, 230, 354, 468, 571, 885, 908, 1486, 1674, 2250, 3045, 3586, 5418, 5322, 8186, 9560, 12234, 16341, 17976, 27912, 26970, 38435, 46383, 57024, 76794, 80805, 125205, 116376, 165914, 201580, 232352, 322980, 332388, 508710, 456154, 645661, 800495, 886018, 1241190, 1226382, 1916682, 1693454, 2290704, 2935492
Offset: 1

Views

Author

Paul D. Hanna, Oct 06 2015

Keywords

Comments

The integer floor values, [n/r^2] and [n/r^3] where r^2 + r^3 = 1, form Beatty sequences and thus together contain all the positive integers without repetition.
Here r = 6 / ( (108 + 12*sqrt(69))^(1/3) + (108 - 12*sqrt(69))^(1/3) ) = 0.75487766624669276.... satisfies r^2 + r^3 = 1.
Not equal to A090845.
What is the rate of growth of this sequence?

Examples

			G.f.: A(x) = x + x^2 + 2*x^3 + 3*x^4 + 5*x^5 + 9*x^6 + 10*x^7 + 20*x^8 + 22*x^9 + 40*x^10 + 51*x^11 + 67*x^12 + 114*x^13 + 126*x^14 + 203*x^15 +...
where the terms are formed from the union of coefficients in A(x)^2 and A(x)^3.
The coefficients of A(x)^2 begin:
A^2 = [1, 2, 5, 10, 20, 40, 67, 126, 203, 354, 571, 908, 1486, 2250, 3586, 5322, 8186, 12234, 17976, 26970, 38435, 57024, 80805, 116376, 165914, 232352,...]
and form the terms of this sequence at positions [n/r^2] for n>=1:
{[n/r^2]} = [1, 3, 5, 7, 8, 10, 12, 14, 15, 17, 19, 21, 22, 24, 26, 28, 29, 31, 33, 35, 36, 38, 40, 42, 43, 45, 47, 49, 50, 52, 54, 56, 57, 59, ...].
The coefficients of A(x)^3 begin:
A^3 = [1, 3, 9, 22, 51, 114, 230, 468, 885, 1674, 3045, 5418, 9560, 16341, 27912, 46383, 76794, 125205, 201580, 322980, 508710, 800495, 1241190, ...]
and form the terms of this sequence at positions [n/r^3] for n>=1:
{[n/r^3]} = [2, 4, 6, 9, 11, 13, 16, 18, 20, 23, 25, 27, 30, 32, 34, 37, 39, 41, 44, 46, 48, 51, 53, 55, 58, 60, 62, 65, 67, 69, 72, ...].

Links

Crossrefs

Cf. A075778.

Programs

  • PARI
    {a(n) = local(A=vector(n+1),B=A,C=A,r=6/((108+12*sqrt(69))^(1/3)+(108-12*sqrt(69))^(1/3))); A[1]=1;A[2]=1;
for(i=1,ceil(log(#A)/log(1/r)),
B=vector(floor(#A/r^2));for(n=1,#A,m=floor(n/r^2);if(m<=#B,B[m]=Vec(Ser(A)^2)[n]));
C=vector(floor(#A/r^3));for(n=1,#A,m=floor(n/r^3);if(m<=#C,C[m]=Vec(Ser(A)^3)[n]));
A=vector(#A,n,if(C[n]==0,B[n],C[n]));); A[n]}
for(n=1,80,print1(a(n),", "))

A090847 Let A denote the sequence; then A is equal to the union of the self-convolutions A^2 and A^4, with terms in ascending order by size, where a(0)=1.

Original entry on oeis.org

1, 1, 2, 4, 5, 12, 14, 22, 44, 50, 88, 117, 160, 308, 309, 508, 740, 912, 1518, 1700, 2470, 3822, 4280, 6606, 8164, 10764, 17158, 17204, 26276, 35020, 42238, 63260, 69664, 97028, 136920, 149924, 219665, 262376, 335600, 493344, 496312, 724942, 925277
Offset: 0

Views

Author

Paul D. Hanna, Dec 09 2003

Keywords

Comments

The occurrences of the terms of A^4 in A is given by A090848. Given A(m)=A^4(n), then what is the limit m/n as n grows? Example: at n=2000, m/n=3202/2000=2.616, at n=3000, m/n=7849/3000=2.6163...

Examples

			A={1,1,2,4,5,12,14,22,44,50,88,117,...} since A is the sorted union of:
A^2={1,2,5,12,22,50,88,160,309,508,912,1518,2470,4280,6606,10764,...} and
A^4={1,4,14,44,117,308,740,1700,3822,8164,17158,35020,69664,136920,...}.

Crossrefs

Cf. A090845, A090848.
Showing 1-6 of 6 results.

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