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.

Previous Showing 21-29 of 29 results.

A353025 Terms of A352991 which are perfect powers.

Original entry on oeis.org

1, 13527684, 34857216, 65318724, 73256481, 81432576, 139854276, 152843769, 157326849, 215384976, 245893761, 254817369, 326597184, 361874529, 375468129, 382945761, 385297641, 412739856, 523814769, 529874361, 537219684, 549386721, 587432169, 589324176, 597362481, 615387249, 627953481, 653927184
Offset: 1

Views

Author

Marco Ripà, Apr 17 2022

Keywords

Comments

It appears that all terms are terms of A062503.
We note that a(n)=A352329(n) up to a(36)=A352329(36)=923187456, while the mentioned match does not hold starting from a(37)=14102987536 (since A352329(37)=1234608769).
There are no perfect powers among terms t which are permutations of 123_...(m - 1)_m for m == {2, 3, 5, 6} (mod 9). This is since 10 == 1 (mod 9) and also (1 + 0) == 1 (mod 9), so digit position has no effect. Hence, t == A134804(m) (mod 9). Now, if m is such that A134804(m) = {3, 6}, there is a lone factor of 3, which is not a perfect power (indeed).
Therefore, all terms are necessarily congruent modulo 9 to 0 or 1 (see Marco Ripà link).
All terms up to 10^34 are squares (in particular, there are 67 squares with no more than 17 digits). - Aldo Roberto Pessolano, May 12 2022

Examples

			75910168324 is a term since 75910168324 = 275518^2.

Links

Crossrefs

Cf. A001292, A058183, A062503, A134804, A181129, A352329, A352991.

Programs

  • Mathematica
    z = 1; Do[r = Range[k];
n = ToExpression[StringJoin[ToString[#] & /@ r]];
If[And[Mod[n, 9] != 3, Mod[n, 9] != 6], d = DigitCount[n];
  s = IntegerPart[Sqrt[10^(IntegerLength[n] - 1)]];
  f = IntegerPart[Sqrt[10^(IntegerLength[n])]];
  Do[y = x^2;
   If[DigitCount[y] == d, c = True;
    Do[If[Not[StringContainsQ[ToString[y], ToString[i]]],
      c = False], {i, 10, k}]; If[c, Print[z, " ", y]; z++]], {x, s,
    f}]], {k, 1, 10}] (* Aldo Roberto Pessolano, May 12 2022 *)

Formula

Digit sum of a(n) is always congruent to 0 or 1 modulo 9.
a(n) = m^2, where the integer m := m(n) is not a perfect power itself (conjectured).

A180346 Primes that divide every circular permutation of the digits of at least one number of the form 123...(n-1)(n) (see A007908), where n is 3 digits long (that is, for some n in the range 99

Original entry on oeis.org

3, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 53, 61, 67, 73, 83, 97, 101, 107, 127, 163, 211, 271, 277, 1009, 18973
Offset: 1

Views

Author

Marco Ripà, Jan 22 2011

Keywords

Comments

Every a(i) divides at least 192 permutations of the digits of an element belonging to [A007908]. Skipping the trivial case a(1)=3, the most recurring elements are a(2)=7 and a(10)=37. The occurrences in our 1386450 terms set are the following [A181373]:
a(2) | 7 ⇒ n=100+14*v (v=0,1,2,...,64)
a(3) | 11 ⇒ n=106+22*v (v=0,1,2,...,40)
a(4) | 13 ⇒ n=120+26*v (v=0,1,2,...,33)
a(5) | 17 ⇒ n=196+272*v (v=0,1,2)
a(6) | 19 ⇒ n=102+114*v (v=0,1,2,3,4,5,6,7)
a(7) | 23 ⇒ n=542
a(8) | 29 ⇒ n=400
a(9) | 31 ⇒ n=181+155*v (v=0,1,2,3,4,5)
a(10)| 37 ⇒ n=123+d(v),
(where d(v)=0,12,25,12,25,12,25... for v=0,1,2,3,...,47)
a(11) | 41 ⇒ n=216+205*v (v=0,1,2,3)
a(12) | 43 ⇒ n=372+301*v (v=0,1,2)
a(13) | 53 ⇒ n=127+689*v (v=0,1)
a(14) | 61 ⇒ n=616
a(15) | 67 ⇒ n=399
a(16) | 73 ⇒ n=196+584*v (v=0,1)
a(17) | 83 ⇒ n=118
a(18) | 97 ⇒ n=516
a(19) | 101 ⇒ n=416+404*v (v=0,1)
a(20) | 107 ⇒ n=884
a(21) | 127 ⇒ n=106
a(22) | 163 ⇒ n=576
a(23) | 211 ⇒ n=306
a(24) | 271 ⇒ n=936
a(25) | 277 ⇒ n=174
a(26) | 1009 ⇒ n=960
a(27) | 18973 ⇒ n=903
N.B.
Every coefficient of "v" is a multiple of i. This is a general property of [A007908], valid for an arbitrary fixed digits interval of the parameter "n" (10^k-1
a(28) >= prime(10^6) if it exists. - Chai Wah Wu, Nov 12 2015
Primes p such that p divides both A007908(m) and 10^A058183(m)-1 for some 99Chai Wah Wu, Oct 07 2023
a(28) > prime(2.3316*10^9) if it exists. Conjecture: 18973 is the last term. - Chai Wah Wu, Oct 09 2023

References

  • Vassilev-Missana and K. Atanassov, “Some Smarandache problems”, Hexis, 2004.

Links

Crossrefs

Cf. A007908, A058183, A181373.

Programs

  • PARI
    isA180346(p)={ isprime(p) & p!=2 & p!=5 & for(n=100,999, my(S=eval(concat(vector(n,i,Str(i)))),L=#Str(S)-1); S%p & next; for(k=1,L, (S=[1,10^L]*divrem(S,10))%p & next(2));return(n)) }  /* returns the least corresponding n or 0 if not in this sequence */ \\ M. F. Hasler, Jan 23 2011
  • Python
    from itertools import islice
from sympy import nextprime
def A180346_gen(startvalue=1): # generator of terms >= startvalue
    p = max(startvalue-1,0)
    while (p:=nextprime(p)):
        c, q, a, b = 0, 1, 10, 10
        for m in range(1,1000):
            if m >= b:
                a = 10*a%p
                b *= 10
            c = (c*a + m) % p
            q = q*a % p
            if m>99 and not (c or (q-1)%p):
                yield p
                break
A180346_list = list(islice(A180346_gen(),20)) # Chai Wah Wu, Oct 07 2023

Formula

For n<10 the only a(i) is 3. If 9

A278814 a(n) = ceiling(sqrt(3n+1)).

Original entry on oeis.org

1, 2, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18
Offset: 0

Views

Author

Mohammad K. Azarian, Nov 28 2016

Keywords

Crossrefs

Cf. A002162, A016777, A016789, A016933, A017569, A032765, A058183, A131033, A007494, A051536, A007559.

Programs

  • Derive
    PROG(y := [], n := 100, LOOP(IF(n = -1, RETURN y), y := ADJOIN(CEILING(SQRT(1 + 3·n)), y), n := n - 1))
  • Maple
    seq(ceil(sqrt(3*k+1)), k=0..100); # Robert Israel, Nov 28 2016
  • Mathematica
    Table[Ceiling[Sqrt[3n+1]],{n,0,100}]
  • PARI
    a(n)=sqrtint(3*n)+1 \\ Charles R Greathouse IV, Nov 29 2016
  • Python
    from math import isqrt
def A278814(n): return 1+isqrt(3*n) # Chai Wah Wu, Jul 28 2022

Formula

a(n) = ceiling(sqrt(3n+1)).
From Robert Israel, Nov 28 2016: (Start)
G.f.: (1-x)^(-1)*Sum_{k>=0} (x^(3*k^2)+x^(3*k^2+2*k+1)+x^(3*k^2+4*k+2)).
a(n+1) = a(n)+1 if n is in A032765, otherwise a(n+1) = a(n). (End)
Sum_{n>=0} (-1)^n/a(n) = log(2) (A002162). - Amiram Eldar, Jun 18 2025

A333183 Number of digits in concatenation of first n positive even integers.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 97, 100, 103, 106, 109, 112, 115, 118, 121, 124, 127, 130, 133, 136, 139, 142, 145, 148, 151, 154
Offset: 1

Views

Author

Alexander Goebel, Mar 10 2020

Keywords

Comments

Connected with A019520 and A038396, similar to how A058183 applies to both A007908 and A000422 to count the digits in them, as the order of the digits does not matter (2468 returns the same result as 8642).

Examples

			For example, a(5) = 6 because 246810 (the concatenation of the first five positive even integers) has six digits.

Links

Crossrefs

Cf. A058183, A007908, A000422, A019520, A038396.

Programs

Formula

a(n) = A058183(n) - Sum_{1..A058183(n)} A000035(A058183(n)).
a(n) = Sum_{i=1..n} (1+floor(log_10(2*i))). - Robert Israel, Apr 05 2020

A362680 a(n) is the number of decimal digits in A173426(n).

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 13, 15, 17, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 104, 108, 112, 116, 120, 124, 128, 132, 136, 140, 144, 148, 152, 156, 160, 164, 168, 172, 176, 180, 184, 188, 192, 196, 200, 204, 208, 212, 216, 220, 224, 228, 232
Offset: 1

Views

Author

David Cleaver, Apr 29 2023

Keywords

Examples

			a(12)=28 since 1234567891011121110987654321 has 28 digits.

Links

Crossrefs

Cf. A173426, A058183, A055642.

Programs

  • Mathematica
    a[n_]:=IntegerLength[FromDigits[Flatten[IntegerDigits/@Join[Range[n], Reverse[Range[n-1]]]]]]; Array[a,63] (* Stefano Spezia, Apr 16 2025 *)
  • PARI
    a(n)={my(t=logint(n,10)+1); 2*n*t-2*(10^t-1)/9+t}
  • Python
    def a(n): return ((n*(t:=len(str(n)))-(10**t-1)//9)<<1) + t
print([a(n) for n in range(1, 64)]) # Michael S. Branicky, May 02 2023

Formula

a(n) = A058183(n) + A058183(n-1), for n >= 2.
a(n) = A055642(A173426(n)).
a(n) = 2*A058183(n) - A055642(n).

A019524 Duplicate terms of A007908.

Original entry on oeis.org

11, 1212, 123123, 12341234, 1234512345, 123456123456, 12345671234567, 1234567812345678, 123456789123456789, 1234567891012345678910, 12345678910111234567891011, 123456789101112123456789101112
Offset: 1

Views

Author

R. Muller

Keywords

References

  • F. Iacobescu, Smarandache Partition Type and Other Sequences, Bull. Pure Appl. Sciences, Vol. 16E, No. 2 (1997), pp. 237-240.

Links

Formula

a(n) = A007908(n)*(1 + 10^A058183(n)) = (n + a(n - 1)*10^L(n)/(1 + 10^(n*L(n - 1) - (10^L(n - 1) - 1)/9)))*(1 + 10^((n + 1)*L(n) - (10^L(n) - 1)/9)) where L(n) = floor(log_10(10n)). - Henry Bottomley, Nov 17 2000 (may need to be adapted for change in offset)

A071423 a(n) = a(n-1) + number of decimal digits of 2^n. Number of decimal digits of concatenation of first n powers of 2.

Original entry on oeis.org

1, 2, 3, 5, 7, 9, 12, 15, 18, 22, 26, 30, 34, 39, 44, 49, 55, 61, 67, 74, 81, 88, 95, 103, 111, 119, 128, 137, 146, 156, 166, 176, 186, 197, 208, 219, 231, 243, 255, 268, 281, 294, 307, 321, 335, 349, 364, 379, 394, 410, 426, 442, 458, 475, 492, 509, 527, 545
Offset: 1

Views

Author

Labos Elemer, May 27 2002

Keywords

Crossrefs

Cf. A058183.

Programs

  • Mathematica
    Do[s=s+Length[IntegerDigits[2^n]]; Print[s], {n, 1, 128}]
nxt[{n_,a_}]:={n+1,a+IntegerLength[2^(n+1)]}; NestList[nxt,{1,1},60][[All,2]] (* Harvey P. Dale, Nov 11 2022 *)

Formula

a(n) = a(n-1)+A034887(n). [R. J. Mathar, Sep 11 2009]
a(n) = 0.5 log 2/log 10 * n^2 + O(n)

Extensions

An incorrect g.f. was deleted by N. J. A. Sloane, Sep 13 2009
Formula from Charles R Greathouse IV, Apr 28 2010

A071424 a(n) = a(n-1) + number of decimal digits of n!. Number of decimal digits of concatenation of first n factorials.

Original entry on oeis.org

1, 2, 3, 5, 8, 11, 15, 20, 26, 33, 41, 50, 60, 71, 84, 98, 113, 129, 147, 166, 186, 208, 231, 255, 281, 308, 337, 367, 398, 431, 465, 501, 538, 577, 618, 660, 704, 749, 796, 844, 894, 946, 999, 1054, 1111, 1169, 1229, 1291, 1354, 1419, 1486, 1554, 1624
Offset: 1

Views

Author

Labos Elemer, May 27 2002

Keywords

Crossrefs

Cf. A058183.

Programs

  • Mathematica
    Do[s=s+Length[IntegerDigits[n! ]]; Print[s], {n, 1, 128}]

A316492 Numbers k such that the average digit in the concatenation of the numbers from 1 through k is an integer.

Original entry on oeis.org

1, 3, 5, 7, 9, 122, 576, 1422, 1876, 4122, 4576
Offset: 1

Views

Author

Jon E. Schoenfield, Aug 11 2018

Keywords

Comments

Equivalently, numbers k such that A058183(k) divides A037123(k).
4576 is the final term; 4 < A037123(k)/A058183(k) < 5 for all k > 4576.

Examples

			9 is a term because the average digit in 123456789 is (1+2+3+4+5+6+7+8+9)/9 = 45/9 = 5 (an integer).
122 is a term because 12345789101112..119120121122 has digit sum 1032 and digit count 258, and 1032/258 = 4 (an integer).

Crossrefs

Cf. A033713, A034967, A037123, A058183, A114136.

Programs

  • Mathematica
    Flatten@ Position[ Divide @@@ Transpose[ Accumulate /@ {Total /@ #, Length /@ #} &@ IntegerDigits@ Range@ 5000], Integer] (* _Giovanni Resta, Aug 12 2018 *)
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