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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A218739 a(n) = (36^n - 1)/35.

Original entry on oeis.org

0, 1, 37, 1333, 47989, 1727605, 62193781, 2238976117, 80603140213, 2901713047669, 104461669716085, 3760620109779061, 135382323952046197, 4873763662273663093, 175455491841851871349, 6316397706306667368565, 227390317427040025268341, 8186051427373440909660277
Offset: 0

Views

Author

M. F. Hasler, Nov 04 2012

Keywords

Comments

Partial sums of powers of 36 (A009980).

Crossrefs

Programs

Formula

From Vincenzo Librandi, Nov 07 2012: (Start)
G.f.: x/((1 - x)*(1 - 36*x)).
a(n) = 37*a(n-1) - 36*a(n-2).
a(n) = floor(36^n/35). (End)
E.g.f.: exp(x)*(exp(35*x) - 1)/35. - Stefano Spezia, Mar 28 2023

A218741 a(n) = (38^n - 1)/37.

Original entry on oeis.org

0, 1, 39, 1483, 56355, 2141491, 81376659, 3092313043, 117507895635, 4465300034131, 169681401296979, 6447893249285203, 245019943472837715, 9310757851967833171, 353808798374777660499, 13444734338241551098963, 510899904853178941760595, 19414196384420799786902611
Offset: 0

Views

Author

M. F. Hasler, Nov 04 2012

Keywords

Comments

Partial sums of powers of 38 (A009982).

Crossrefs

Programs

Formula

From Vincenzo Librandi, Nov 07 2012: (Start)
G.f.: x/((1-x)*(1-38*x)).
a(n) = 39*a(n-1) - 38*a(n-2).
a(n) = floor(38^n/37). (End)
E.g.f.: exp(x)*(exp(37*x) - 1)/37. - Elmo R. Oliveira, Aug 29 2024

A218742 a(n) = (39^n - 1)/38.

Original entry on oeis.org

0, 1, 40, 1561, 60880, 2374321, 92598520, 3611342281, 140842348960, 5492851609441, 214221212768200, 8354627297959801, 325830464620432240, 12707388120196857361, 495588136687677437080, 19327937330819420046121, 753789555901957381798720, 29397792680176337890150081
Offset: 0

Views

Author

M. F. Hasler, Nov 04 2012

Keywords

Comments

Partial sums of powers of 39 (A009983).

Crossrefs

Programs

Formula

a(n) = floor(39^n/38).
From Vincenzo Librandi, Nov 07 2012: (Start)
G.f.: x/((1-x)*(1-39*x)).
a(n) = 40*a(n-1) - 39*a(n-2). (End)
E.g.f.: exp(20*x)*sinh(19*x)/19. - Elmo R. Oliveira, Aug 29 2024

A218747 a(n) = (44^n - 1)/43.

Original entry on oeis.org

0, 1, 45, 1981, 87165, 3835261, 168751485, 7425065341, 326702875005, 14374926500221, 632496766009725, 27829857704427901, 1224513738994827645, 53878604515772416381, 2370658598693986320765, 104308978342535398113661, 4589595047071557517001085, 201942182071148530748047741
Offset: 0

Views

Author

M. F. Hasler, Nov 04 2012

Keywords

Comments

Partial sums of powers of 44 (A009988).

Crossrefs

Programs

Formula

From Vincenzo Librandi, Nov 07 2012: (Start)
G.f.: x/((1-x)*(1-44*x)).
a(n) = 45*a(n-1) - 44*a(n-2).
a(n) = floor(44^n/43). (End)
E.g.f.: exp(x)*(exp(43*x) - 1)/43. - Elmo R. Oliveira, Aug 29 2024

A218748 a(n) = (45^n - 1)/44.

Original entry on oeis.org

0, 1, 46, 2071, 93196, 4193821, 188721946, 8492487571, 382161940696, 17197287331321, 773877929909446, 34824506845925071, 1567102808066628196, 70519626362998268821, 3173383186334922096946, 142802243385071494362571, 6426100952328217246315696
Offset: 0

Views

Author

M. F. Hasler, Nov 04 2012

Keywords

Comments

Partial sums of powers of 45 (A009989).

Crossrefs

Programs

Formula

G.f.: x/((1-x)*(1-45*x)). - Vincenzo Librandi, Nov 08 2012
a(n) = 46*a(n-1) - 45*a(n-2) with a(0)=0, a(1)=1. - Vincenzo Librandi, Nov 08 2012
a(n) = 45*a(n-1) + 1 with a(0)=0. - Vincenzo Librandi, Nov 08 2012
a(n) = floor(45^n/44). - Vincenzo Librandi, Nov 08 2012
E.g.f.: exp(23*x)*sinh(22*x)/22. - Elmo R. Oliveira, Aug 27 2024

A218749 a(n) = (46^n - 1)/45.

Original entry on oeis.org

0, 1, 47, 2163, 99499, 4576955, 210539931, 9684836827, 445502494043, 20493114725979, 942683277395035, 43363430760171611, 1994717814967894107, 91757019488523128923, 4220822896472063930459, 194157853237714940801115, 8931261248934887276851291, 410838017451004814735159387
Offset: 0

Views

Author

M. F. Hasler, Nov 04 2012

Keywords

Comments

Partial sums of powers of 46 (A009990).

Crossrefs

Programs

Formula

From Vincenzo Librandi, Nov 08 2012: (Start)
G.f.: x/((1-x)*(1-46*x)).
a(n) = 47*a(n-1) - 46*a(n-2) with a(0)=0, a(1)=1.
a(n) = 46*a(n-1) + 1 with a(0)=0.
a(n) = floor(46^n/45). (End)
E.g.f.: exp(x)*(exp(45*x) - 1)/45. - Elmo R. Oliveira, Aug 29 2024

A218751 a(n) = (48^n - 1)/47.

Original entry on oeis.org

0, 1, 49, 2353, 112945, 5421361, 260225329, 12490815793, 599559158065, 28778839587121, 1381384300181809, 66306446408726833, 3182709427618887985, 152770052525706623281, 7332962521233917917489, 351982201019228060039473, 16895145648922946881894705, 810966991148301450330945841
Offset: 0

Views

Author

M. F. Hasler, Nov 04 2012

Keywords

Comments

Partial sums of powers of 48 (A009992).

Crossrefs

Programs

Formula

a(n) = floor(48^n/47).
From Vincenzo Librandi, Nov 08 2012: (Start)
G.f.: x/((1-x)*(1-48*x)).
a(n) = 49*a(n-1) - 48*a(n-2) with a(0)=0, a(1)=1.
a(n) = 48*a(n-1) + 1 with a(0)=0. (End)
E.g.f.: exp(x)*(exp(47*x) - 1)/47. - Elmo R. Oliveira, Aug 29 2024

A330135 a(n) = ((10^(n+1))^4 - 1)/9999 for n >= 0.

Original entry on oeis.org

1, 10001, 100010001, 1000100010001, 10001000100010001, 100010001000100010001, 1000100010001000100010001, 10001000100010001000100010001, 100010001000100010001000100010001
Offset: 0

Views

Author

Bernard Schott, Dec 02 2019

Keywords

Comments

This sequence was the subject of the 6th problem of the 15th British Mathematical Olympiad in 1979 (see the link BMO).
There are no prime numbers in this infinite sequence. Why?
a(0) = 1 and a(1) = 10001 = 73 * 137;
if n even = 2*k, k >= 1, then A094028(n) divides a(n);
if n odd = 2*k+1, k >= 1, then a(k) divides a(n).

Examples

			a(2) = ((10^3)^4 - 1)/9999 = 100010001 = 10101 * 9901 where 10101 = A094028(2).
a(3) = ((10^4)^4 - 1)/9999 = 1000100010001 = 10001 * 100000001 where 10001 = a(1).
From _Omar E. Pol_, Dec 04 2019: (Start)
Illustration of initial terms:
                  1;
                10001;
              100010001;
            1000100010001;
          10001000100010001;
        100010001000100010001;
      1000100010001000100010001;
    10001000100010001000100010001;
  100010001000100010001000100010001;
...
(End)
		

References

  • A. Gardiner, The Mathematical Olympiad Handbook: An Introduction to Problem Solving, Oxford University Press, 1997, reprinted 2011, Pb 6 pp. 68 and 201 (1979).

Crossrefs

Cf. A000533 (1000...0001), A094028 (10101...101), A261544 (1001001...1001).
Cf. A131865 (similar, with 2^(n+1)).

Programs

  • Maple
    A: = seq((10^(4*n+4)-1)/9999, n=1..4);
  • Mathematica
    Table[((10^(n+1))^4 - 1)/9999, {n, 0, 8}] (* Amiram Eldar, Dec 04 2019 *)
  • PARI
    Vec(1 / ((1 - x)*(1 - 10000*x)) + O(x^11)) \\ Colin Barker, Dec 05 2019

Formula

a(n) = (10^(4*n+4) - 1)/9999 for n >= 0.
G.f.: 1 / ((1 - x)*(1 - 10000*x)). - Colin Barker, Dec 05 2019

A325911 Screaming numbers in base 16: numbers whose hexadecimal representation is AAAAAAA...

Original entry on oeis.org

10, 170, 2730, 43690, 699050, 11184810, 178956970, 2863311530, 45812984490, 733007751850, 11728124029610, 187649984473770, 3002399751580330, 48038396025285290, 768614336404564650, 12297829382473034410, 196765270119568550570, 3148244321913096809130
Offset: 1

Views

Author

Eliora Ben-Gurion, Sep 08 2019

Keywords

Comments

In any base b > 10, we may express ten as a digit by using the letter A.

Examples

			a(10) = 733007751850_10 = AAAAAAAAAA_16.
		

Crossrefs

Programs

  • Mathematica
    10Accumulate[16^Range[0, 31]] (* Alonso del Arte, Sep 17 2019 *)
    LinearRecurrence[{17,-16},{10,170},20] (* Harvey P. Dale, Apr 02 2023 *)
  • PARI
    a(n)={10*(16^n-1)/15} \\ Andrew Howroyd, Sep 08 2019
    
  • PARI
    Vec(10*x / ((1 - x)*(1 - 16*x)) + O(x^20)) \\ Colin Barker, Sep 16 2019
  • Python
    a = 10
    while a:
        a = a*16+10
        print(a)
    
  • Python
    def a(n): return int("A"*n, 16)
    print([a(n) for n in range(1, 19)]) # Michael S. Branicky, Jan 17 2022
    

Formula

a(n) = Sum_{i=0..n} 10*16^(i).
a(n) = A131865(n-1)*10.
a(n) = 10*(16^n-1)/15. - Andrew Howroyd, Sep 08 2019
From Colin Barker, Sep 16 2019: (Start)
G.f.: 10*x / ((1 - x)*(1 - 16*x)).
a(n) = 17*a(n-1) - 16*a(n-2) for n>2.
(End)
E.g.f.: (2/3)*exp(x)*(-1 + exp(15*x)). - Stefano Spezia, Sep 17 2019

A352782 The binary expansion of a(n) encodes the runs of consecutive 1's in the binary expansion of n (see Comments section for precise definition).

Original entry on oeis.org

0, 1, 2, 4, 8, 3, 16, 32, 64, 5, 6, 12, 128, 9, 256, 512, 1024, 17, 10, 20, 24, 7, 48, 96, 2048, 33, 18, 36, 4096, 65, 8192, 16384, 32768, 129, 34, 68, 40, 11, 80, 160, 192, 13, 14, 28, 384, 25, 768, 1536, 65536, 257, 66, 132, 72, 19, 144, 288, 131072, 513
Offset: 0

Views

Author

Rémy Sigrist, Apr 02 2022

Keywords

Comments

For any nonnegative integer n:
- the binary expansion of n can be uniquely expressed as the concatenation of k = A069010(n) positive terms of A023758 separated by 0's:
n = A023758(m_k+1) | 0 | A023758(m_{k-1}+1) | 0 | ... | 0 | A023758(m_1+1)
(where | denotes binary concatenation)
- a(n) = ( Sum_{i = 1..k} 2^Sum_{j = 1..i} m_j ) / 2.
This sequence is a permutation of the nonnegative integers, with inverse A352783.

Examples

			For n = 89:
- the binary expansion of 89 is "1011001",
- "1011001" = "1" | 0 | "110" | 0 | "1"
            = A023758(1+1) | 0 | A023758(5+1) | 0 | A023758(1+1)
- so 2*a(89) = 2^(1+5+1) + 2^(5+1) + 2^1 = 194,
- and a(89) = 97.
		

Crossrefs

Programs

  • PARI
    a(n) = { my (v=0, s=-1, z, o, i); while (n, n\=2^z=valuation(n,2); n\=2^o=valuation(n+1,2); n\=2; i=(o+z)*(o+z-1)/2 + o; v+=2^s+=i); v }

Formula

a(4*n+1) = 2*a(n)+1.
A000120(a(n)) = A069010(n).
a(A023758(k+1)) = 2^k for any k >= 0.
a(2^k) = A006125(k+1) for any k >= 0.
a(2^k-1) = A036442(k+1) for any k >= 0.
a(n) = n iff n = 0 or n belongs to A131865 or n/2 belongs to A131865.
Previous Showing 41-50 of 51 results. Next