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.

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A341373 G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies: [Sum_{n>=0} x^n/(1 - x^(n+1))]^2 = Sum_{n>=0} a(n)*x^n/(1 - x^(n+1))^2.

Original entry on oeis.org

1, 2, 5, 6, 15, 6, 30, 16, 34, 14, 69, 12, 95, 20, 51, 46, 152, 8, 179, 34, 90, 34, 253, 26, 210, 46, 174, 52, 371, -30, 402, 128, 179, 36, 254, 40, 527, 78, 225, 58, 647, -56, 673, 140, 178, 38, 813, 46, 600, 32, 334, 104, 963, -24, 467, 180, 381, 26, 1169, -10, 1119, 120, 318, 236, 649
Offset: 0

Views

Author

Paul D. Hanna, Feb 11 2021

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 5*x^2 + 6*x^3 + 15*x^4 + 6*x^5 + 30*x^6 + 16*x^7 + 34*x^8 + 14*x^9 + 69*x^10 + 12*x^11 + 95*x^12 + 20*x^13 + 51*x^14 + 46*x^15 + ...
such that
D(x)^2 = 1/(1-x)^2 + 2*x/(1-x^2)^2 + 5*x^2/(1-x^3)^2 + 6*x^3/(1-x^4)^2 + 15*x^4/(1-x^5)^2 + 6*x^5/(1-x^6)^2 + 30*x^6/(1-x^7)^2 + ... + a(n)*x^n/(1-x^(n+1))^2 + ...
and
D(x)^2 = A(x) + 2*x*A(x^2) + 3*x^2*A(x^3) + 4*x^3*A(x^4) + 5*x^4*A(x^5) + 6*x^5*A(x^6) + 7*x^6*A(x^7) + ... + (n+1)*x^n*A(x^(n+1)) + ...
where
D(x)^2 = 1 + 4*x + 8*x^2 + 14*x^3 + 20*x^4 + 28*x^5 + 37*x^6 + 44*x^7 + 58*x^8 + 64*x^9 + 80*x^10 + 86*x^11 + 108*x^12 + ... + A055507(n+1)*x^n + ...
D(x) = 1 + 2*x + 2*x^2 + 3*x^3 + 2*x^4 + 4*x^5 + 2*x^6 + 4*x^7 + 3*x^8 + 4*x^9 + 2*x^10 + 6*x^11 + 2*x^12 + ... + A000005(n+1)*x^n + ...

Crossrefs

Cf. A341374, A341375, A055507, A000005.

Programs

  • PARI
    {a(n) = my(A=[1]); for(i=1,n, A=concat(A,0);
A[#A] = polcoeff( sum(n=0,#A, x^n/(1 - x^(n+1) +x*O(x^#A)) )^2 - sum(n=0,#A-1,A[n+1]*x^n/(1 - x^(n+1) + x*O(x^#A))^2 ), #A-1) );A[n+1]}
for(n=0,100,print1(a(n),", "))

Formula

G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies:
(1) [ Sum_{n>=0} x^n/(1 - x^(n+1)) ]^2 = Sum_{n>=0} a(n) * x^n / (1 - x^(n+1))^2.
(2) [ Sum_{n>=0} x^n/(1 - x^(n+1)) ]^2 = Sum_{n>=0} (n+1) * x^n * A( x^(n+1) ).

A338671 a(n) is the number of distinct ways of arranging n identical square tiles into two rectangles.

Original entry on oeis.org

0, 1, 1, 2, 3, 4, 5, 7, 7, 10, 10, 13, 14, 16, 16, 21, 20, 25, 24, 29, 28, 34, 30, 40, 36, 43, 40, 50, 44, 55, 49, 60, 55, 66, 55, 75, 64, 75, 70, 86, 72, 92, 77, 97, 87, 103, 84, 116, 94, 114, 104, 126, 102, 135, 109, 138, 123, 143, 117, 164, 128, 153, 138, 171
Offset: 1

Views

Author

Thomas Oléron Evans, Apr 23 2021

Keywords

Comments

Note that rectangles of size 0 are not accepted (i.e., the tiles may not be formed into a single rectangle).
Finding a(n) is equivalent to counting the positive integer solutions (x,y,z,w) of n = xy + zw such that:
i) x >= y,z,w
ii) z >= w
iii) if x = z then y >= w
These conditions ensure that identical pairs are not counted twice by permuting the values of the variables.

Examples

			a(5) = 3, since there are 3 ways to form 2 rectangles from 5 identical square tiles:
1) 2 X 2  and  1 X 1
2) 3 X 1  and  2 X 1
3) 4 X 1  and  1 X 1
Note that rotation through 90 degrees and/or exchanging the order of the two rectangles in a pair naturally do not create a distinct pair. For example, the pair 2 X 1 and 1 X 3 is not distinct from pair 2 above.

Crossrefs

Cf. A038548, A338664, A055507 (where a(n) is the number of ordered ways to express n+1 as a*b+c*d with 1 <= a,b,c,d <= n).

Programs

  • PARI
    a(n) = {(sum(k=1, n-1, ((numdiv(k)+1)\2)*((numdiv(n-k)+1)\2)) + if(n%2==0, (numdiv(n/2)+1)\2))/2} \\ Andrew Howroyd, Apr 29 2021
  • Python
    import numpy as np
# This sets the number of terms:
nits = 20
# This list will contain the sequence
seq = []
# The indices of the sequence:
for i in range(1,nits + 1):
    # This variable counts the pairs found for each total area i
    count = 0
    # The longest side of either rectangle:
    for a in range(1,i):
        # The other side of the same rectangle:
        for b in np.arange(1,1 + min(a,np.floor(i/a))):
            # Calculate the area of this rectangle and the remaining area:
            area1    = a*b
            rem_area = i - a*b
            # The longer side of the second rectangle:
            for c in np.arange(1,1 + min(a,rem_area)):
                # The shorter side of the second rectangle:
                d = rem_area / c
                # Check that the solution is valid and not double counted:
                if d != int(d) or d > min(a,c) or (a == c and d > b):
                    continue
                # Count the new pair found:
                count += 1
    # Add to the sequence:
    seq.append(count)
for an in seq:
    print(an)

Formula

G.f.: (B(x)^2 + B(x^2))/2 where B(x) is the g.f. of A038548. - Andrew Howroyd, Apr 29 2021

A338810 a(n) = (n!/2) * Sum_{k=1..n-1} d(k)*d(n-k)/(k*(n-k)), where d(n) is the number of divisors of n.

Original entry on oeis.org

0, 1, 6, 28, 170, 988, 7896, 60492, 555264, 5819904, 61776000, 725950080, 9894493440, 137963243520, 1875645434880, 33258387456000, 528975488563200, 9760969019289600, 175565885864140800, 3608256006957772800, 72367669059194880000, 1745463407406243840000
Offset: 1

Views

Author

Seiichi Manyama, Nov 10 2020

Keywords

Crossrefs

Column 2 of A338805.
Cf. A000005, A055507, A059356.

Programs

  • Mathematica
    a[n_] := (n - 1)! * Sum[DivisorSigma[0, k] * DivisorSigma[0, n - k]/k, {k, 1, n - 1} ]; Array[a, 22] (* Amiram Eldar, Nov 10 2020 *)
  • PARI
    {a(n)= n!*sum(k=1, n-1, numdiv(k)*numdiv(n-k)/(k*(n-k)))/2}
  • PARI
    {a(n)= (n-1)!*sum(k=1, n-1, numdiv(k)*numdiv(n-k)/k)}
  • PARI
    {a(n) = my(u='u); n!*polcoef(polcoef(prod(k=1, n, (1-x^k+x*O(x^n))^(-u/k)), n), 2)}

Formula

a(n) = (n-1)! * Sum_{k=1..n-1} d(k)*d(n-k)/k.

A341636 a(n) = Sum_{d|n} phi(d) * tau(d) * tau(n/d).

Original entry on oeis.org

1, 4, 6, 13, 10, 24, 14, 38, 29, 40, 22, 78, 26, 56, 60, 103, 34, 116, 38, 130, 84, 88, 46, 228, 79, 104, 124, 182, 58, 240, 62, 264, 132, 136, 140, 377, 74, 152, 156, 380, 82, 336, 86, 286, 290, 184, 94, 618, 153, 316, 204, 338, 106, 496, 220, 532, 228, 232, 118, 780, 122, 248
Offset: 1

Views

Author

Ilya Gutkovskiy, Feb 16 2021

Keywords

Comments

Inverse Moebius transform of A062949.

Links

Crossrefs

Cf. A000005, A000010, A000203, A007426, A055507, A062355, A062949.

Programs

  • Mathematica
    Table[Sum[EulerPhi[d] DivisorSigma[0, d] DivisorSigma[0, n/d], {d, Divisors[n]}], {n, 62}]
Table[Sum[DivisorSigma[0, GCD[n, k]] DivisorSigma[0, n/GCD[n, k]], {k, n}], {n, 62}]
f[p_, e_] := (p + 1 + e*(p - 1) + p^(e + 1)*(e*(p - 1) + p - 3))/(p - 1)^2; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 15 2023 *)
  • PARI
    a(n) = sumdiv(n, d, eulerphi(d)*numdiv(d)*numdiv(n/d)); \\ Michel Marcus, Feb 17 2021

Formula

a(n) = Sum_{k=1..n} tau(gcd(n,k)) * tau(n/gcd(n,k)).
a(n) = Sum_{d|n} A062949(d).
Multiplicative with a(p^e) = (p + 1 + e*(p-1) + p^(e+1)*(e*(p-1)+p-3))/(p-1)^2. - Amiram Eldar, Sep 15 2023

A374973 a(n) = Sum_{k=1..n-1} tau(k) * sigma_2(n-k).

Original entry on oeis.org

0, 1, 7, 22, 54, 105, 188, 307, 459, 690, 937, 1307, 1680, 2260, 2740, 3588, 4221, 5402, 6163, 7714, 8694, 10723, 11758, 14449, 15574, 18884, 20320, 24228, 25626, 30768, 32038, 37985, 39826, 46515, 47898, 56877, 57754, 67433, 69450, 80062, 81103, 95034, 94941
Offset: 1

Views

Author

Seiichi Manyama, Jul 26 2024

Keywords

Comments

Convolution of tau with sigma_2.

Crossrefs

Cf. A000005, A001157, A055507, A191831.

Programs

  • PARI
    a(n) = sum(k=1, n-1, sigma(k, 0)*sigma(n-k, 2));
  • PARI
    my(N=50, x='x+O('x^N)); concat(0, Vec(sum(k=1, N, x^k/(1-x^k))*sum(k=1, N, k^2*x^k/(1-x^k))))

Formula

G.f.: ( Sum_{k>=1} x^k/(1 - x^k) ) * ( Sum_{k>=1} k^2 * x^k/(1 - x^k) ).

A375002 a(n) = Sum_{i+j+k+m=n, i,j,k,m >= 1} tau(i) * tau(j) * tau(k) * tau(m).

Original entry on oeis.org

0, 0, 0, 1, 8, 32, 92, 216, 440, 814, 1392, 2244, 3452, 5096, 7292, 10129, 13760, 18284, 23868, 30662, 38820, 48556, 59948, 73424, 88796, 106886, 127052, 150732, 176560, 206920, 239344, 277616, 317516, 365034, 413508, 471637, 529712, 600076, 668708, 753070, 833408
Offset: 1

Views

Author

Seiichi Manyama, Jul 27 2024

Keywords

Comments

4-fold convolution of tau (A000005).

Crossrefs

Column k=4 of A320019.
Cf. A000005, A055507.

Programs

  • PARI
    my(N=50, x='x+O('x^N)); concat([0, 0, 0], Vec(sum(k=1, N, x^k/(1-x^k))^4))

Formula

G.f.: ( Sum_{k>=1} x^k/(1 - x^k) )^4.
a(n) = Sum_{i=1..n-3} A055507(i)*A055507(n-2-i). - Chai Wah Wu, Jul 27 2024
Previous Showing 21-26 of 26 results.

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