A014824 -id:A014824 - cp's OEIS frontend

A178643 Square array read by antidiagonals. Convolution of a(n) = 2*a(n-1) - a(n-2) and 10^n.

1, 10, 2, 100, 19, 4, 1000, 190, 36, 8, 10000, 1900, 361, 68, 16, 100000, 19000, 3610, 686, 128, 32, 1000000, 190000, 36100, 6859, 1304, 240, 64, 10000000, 1900000, 361000, 68590, 13032, 2480, 448, 128, 100000000, 19000000, 3610000, 685900, 130321, 24760, 4720, 832, 256

Offset: 1

Mark Dols, May 31 2010

Diagonals sum up to A014824.

Alternating diagonal sum gives decimal expansion of fraction 1/119 (A021123).

			Array starts:
     1,    2,    4,    8,
    10,   19,   36,   68,
   100,  190,  361,  686,
  1000, 1900, 3610, 6859,

Robin Visser, Table of n, a(n) for n = 1..10000

Cf. A014824, A000129, A021123, A021083, A178511.

def a(n,k):
    T = [[0 for j in range(k+1)] for i in range(n+1)]
    for i in range(n+1): T[i][0] = 10^i
    for j in range(1, k+1):
        T[0][j] = 2^j
        for i in range(1, n+1): T[i][j] = 2*T[i][j-1] - T[i-1][j-1]
    return T[n][k]  # Robin Visser, Aug 09 2023

T(n,k) = 2*T(n,k-1) - T(n-1,k-1) for all n, k > 0, where T(n,0) = 10^n and T(0,k) = 2^k. - Robin Visser, Aug 09 2023

More terms from Robin Visser, Aug 09 2023

A180027 Partial sums of A100706.

Original entry on oeis.org

1, 112, 11223, 1122334, 112233445, 11223344556, 1122334455667, 112233445566778, 11223344556677889, 1122334455667789000, 112233445566778900111, 11223344556677890011222, 1122334455667789001122333, 112233445566778900112233444, 11223344556677890011223344555, 1122334455667789001122334455666

Offset: 0

Mark Dols, Aug 07 2010

nonn

Up to n=8 the digits of a(n) sum up to n^2.
Similar to this, A014824 (1,12,123,1234,...) is a representation of the triangular numbers; (1,1112,1112223,1112223334,...) of the pentagonal numbers;(1,11112,111122223,...) of the hexagonal numbers, and so on. A nice thing about this sequence(s) is that the (represented) value of the integer matches the partial sums of the number of digits in the sequence.
f(n) = 100*f(n-1) + A100706(n) gives a mirrored version of this sequence, and f(n) = 10*f(n-1) + A100706(n) the symmetrical version (A002477).

Cf. A014824, A100706, A002477.

A100706(n) = (10^(2*n + 1) - 1)/9;
a(n) = sum(k=0, n, A100706(k)); \\ Michel Marcus, Mar 12 2023

a(n) = Sum_{k=0..n} A100706(k). - Michel Marcus, Mar 12 2023

More terms and edited by Michel Marcus, Mar 12 2023

A180039 Combinatorial tetrahedral numbers.

Original entry on oeis.org

1, 112, 111223, 1111222334, 111112222333445, 111111222223333444556, 1111111222222333334444555667, 111111112222222333333444445555666778, 111111111222222223333333444444555556666777889

Offset: 1

Mark Dols, Aug 07 2010

These are partial sums in adding base-1 triangular numbers (1 + 111 + 111111 + 1111111111,...).
Third row of:
1,2,3,4,5,...
1,12,123,1234,12345,...
1,112,111223,1111222334,111112222333445,...
1,1112,1111112223,11111111112222223334,11111111111111122222222223333334445,..

			For n=4, a(4)= 1 + 111 + 111111 + 1111111111 = 1111222334.
Combinatorially:
1,12,123,1234,12345,123456,...
1,21,321,4321,54321,654321,...
----------------------------- X
1,112,111223,1111222334.......
For n =(4),a(4)= 4 x 1 + 3 x 2 + 2 x 3 + 1 x 4 = 1111 + 222 + 33 + 4 = 1111222334.

Cf. A000292, A014824, A180027.

Accumulate[Table[FromDigits[PadRight[{},n,1]],{n,Accumulate[Range[10]]}]] (* Harvey P. Dale, Jun 12 2021 *)

a(n) = Sum_{i=1..n} A002275(A000217(i)). - R. J. Mathar, Nov 02 2016

A249181 a(n) = A057137(n)^2 where A057137 = 0,1,12,123,...,123...90,...

Original entry on oeis.org

0, 1, 144, 15129, 1522756, 152399025, 15241383936, 1524155677489, 152415765279684, 15241578750190521, 1524157875019052100, 152415787526596567801, 15241578753153483936144, 1524157875322755800955129, 152415787532374345526722756, 15241578753238669120562399025

Offset: 0

M. F. Hasler, Oct 22 2014

A very common playful operation on pocket calculators is to type as many digits 123... as the display allows and then squaring it by pressing the "X" and then the "=" key. On basic pocket calculators this yields an overflow with the first digits of, e.g., a(8) displayed, viz "E 1524157.6".

			a(3) = 123^2 = 15129.
a(10) = 1234567890^2 = 1524157875019052100.

Index entries for linear recurrences with constant coefficients, signature (100, 0, 0, 0, 0, 0, 0, 0, 0, 10000000001, -1000000000100, 0, 0, 0, 0, 0, 0, 0, 0, -10000000000, 1000000000000).

[(137174210*10^n div 1111111111)^2: n in [0..20]]; // Vincenzo Librandi, Oct 23 2014

print1(t=0);for(i=1,19,t=t*10+i%10;print1(","t*t))

A249181(n)=(137174210*10^n\1111111111)^2

For n<10, a(n) = A014824(n)^2 = floor(10^(n+1)/81-n/9)^2.

a(n) ~ 1.524157875...*10^(2n-2).

A332082 a(n) = Sum_{1 <= m <= n} Sum_{1 <= k <= n+1-m} m*R(k,n+1), where R(k,b) = (b^k - 1)/(b - 1) is the base-b repunit of length k.

Original entry on oeis.org

0, 1, 7, 42, 295, 2675, 31122, 447188, 7661370, 152415765, 3452271185, 87693358654, 2468455488681, 76256200336407, 2564715882332660, 93281313241869480, 3647955866777821668, 152635100350763019705, 6803550294289868214315, 321844061970058547739730, 16103630469426364324556635

Offset: 0

M. F. Hasler, Aug 19 2020

This is the sum of all elements of a triangular matrix (of size n given by the index) where the k-th diagonal is filled with k-repdigits (in base n+1, as to have digits up to n) of increasing length, as in
  ( 1   0    0    . . . . . . . .  0   )
  ( 2  11    0                     :   )
  ( 3  22   111   ˙ · .            :   )
  ( :  33   222   ˙ · .  ˙ · .     :   )
  ( :      ˙ · .  ˙ · .  ˙ · .     0   )
  ( n  . . . . .   3..3  2...2  1....1 )
with numbers written in base n+1.

			a(2) = 2 + 1 + 11[3] = 3 + 4 = 7.
a(3) = 3 + 2 + 22[4] + 1 + 11[4] + 111[4] = 6 + 15 + 21 = 42.
a(4) = 4 + 3 + 33[5] + 2 + 22[5] + 222[5] + 1 + 11[5] + 111[5] + 1111[5] = 10 + 36 + 93 + 156 = 295.

For base 10 repunits and repdigits, cf. A002275 (repunits), A010785 (repdigits) and A014824 (partial sums of repunits).

a[n_] := Sum[(n + 1 - m)*((n + 1)*((n + 1)^m - 1) - m*n)/n^2, {m, 1, n}]; Array[a, 21, 0] (* Amiram Eldar, Aug 24 2020 *)

apply( {A332082(n)=sum(m=1,n,(n+1-m)*((n+1)*((n+1)^m-1)-m*n)\n^2)}, [0..20])

A359846 a(n) = (((5 - (n mod 2))10^(3 + n(9/2) - (n mod 2)*(5/2)))^2 + 2)/81.

Original entry on oeis.org

308642, 1975308642, 308641975308641975308642, 1975308641975308641975308642, 308641975308641975308641975308641975308642, 1975308641975308641975308641975308641975308642, 308641975308641975308641975308641975308641975308641975308642

Offset: 0

Thomas Scheuerle, Jan 15 2023

Also numbers of the form ((d*10^k)^2 + 2)/9^2 that are not squares, where d is a single-digit number.

The square roots of these numbers show runs of equal digits, see the link to Schizophrenic numbers.

			a(0) = 308642 and sqrt(a(0)) = 555.555577777777333333351111110222222271999997013333521... .
a(1) = 1975308642 and sqrt(a(1)) = 44444.4444447222222222213541666666720920138888465033637... .

Paolo Xausa, Table of n, a(n) for n = 0..100
K. S. Brown, Schizophrenic numbers
Mathematics StackExchange, Is there any mathematical reason for this 'digit-repetition-show'?
Wikipedia, Schizophrenic Number
Index entries for linear recurrences with constant coefficients, signature (1,1000000000000000000,-1000000000000000000).

Cf. A014824.

A359846[n_] := (10^(9*n - 5*# + 6)*(# - 5)^2 + 2)/81 & [Mod[n, 2]]; Array[A359846, 10, 0] (* or *)
LinearRecurrence[{1, 10^18, -10^18}, {308642, 1975308642, 308641975308641975308642}, 10] (* Paolo Xausa, Jan 23 2025 *)

a(n) = (((5-(n%2))*10^(3+n*(9/2)-(n%2)*(5/2)))^2+2)/81

G.f.: (2/81)*(1/(1-x)+6249960/(1+1000000000*x)+6250040/(1-1000000000*x)).
a(n) = a(n-1) + 10^18*a(n-2) - 10^18*a(n-3).
a(2*n) = (25*10^(6 + 18*n) + 2)/81.
a(2*n + 1) = (16*10^(10 + 18*n) + 2)/81.
We use in the next formulas a special notation for real numbers where (x) after a digit denotes a run of length x for this digit. Example: 3(4).2(3) is 3333.2222 .
sqrt(a(2*n)) = 5(3+9*n).5(4+9*n)7(8+18*n)3(7+18*n)5(1)1(6+18*n)0(1)2(7+18*n)7(1)1(1)9(5+18*n)7(1)0(1)1(1)3(4+18*n)... .
sqrt(a(2*n+1)) = 4(5+9*n).4(6+9*n)7(1)2(10+18*n)1(1)3(1)5(1)4(1)1(1)6(6+18*n)7(1)2(1)0(1)9(1)2(1)0(1)1(1)3(1)8(4+18*n)... .
sqrt(1/a(2*n)) = 0.0(2+9*n)1(1)7(1)9(6+18*n)2(1)8(1)0(5+18*n)... .
sqrt(1/a(2*n+1)) = 0.0(4+9*n)2(2)4(1)9(8+18*n)8(1)5(1)9(1)3(1)7(1)5(1)0(5+18*n)... .
sqrt(a(2*n)-(2/81)) = 10^(4+9*n)/18.
sqrt(a(2*n+1)-(2/81)) = 10^(7+9*n)/225.

A055528 a(n) = 10*a(n-1)+n^3, a(0)=0.

Original entry on oeis.org

0, 1, 18, 207, 2134, 21465, 214866, 2149003, 21490542, 214906149, 2149062490, 21490626231, 214906264038, 2149062642577, 21490626428514, 214906264288515, 2149062642889246, 21490626428897373, 214906264288979562, 2149062642889802479, 21490626428898032790

Offset: 0

Henry Bottomley, Jul 04 2000

a(n)/10^n converges to 470/2187=0.214906264...

Index entries for linear recurrences with constant coefficients, signature (14,-46,64,-41,10).

Cf. A014824.

RecurrenceTable[{a[0]==0,a[n]==10a[n-1]+n^3},a,{n,20}] (* Harvey P. Dale, Mar 21 2023 *)

concat(0, Vec(-x*(x^2+4*x+1)/((x-1)^4*(10*x-1)) + O(x^100))) \\ Colin Barker, Sep 13 2014

a(n) = (10^n-1)*(470/2187)-n^3/9-n^2*(10/27)-n*(110/243).

G.f.: -x*(x^2+4*x+1) / ((x-1)^4*(10*x-1)). - Colin Barker, Sep 13 2014

Corrected by T. D. Noe, Nov 08 2006

More terms from Colin Barker, Sep 13 2014

A055530 The recurrence b(k) = 10*b(k-1) + k^n with b(0)=0 has b(k)/10^k converging to a(n)/9^(n+1).

Original entry on oeis.org

1, 10, 110, 1410, 22110, 428610, 10027710, 274463010, 8585407710, 302029998210, 11804909261310, 507547187120610, 23805911748929310, 1209638912316543810, 66192799008847310910, 3880867089138927234210, 242703222549879015746910

Offset: 0

Henry Bottomley, Jul 04 2000

Alex Walker, On the Growth of Sequences, 2007

Cf. A002275, A014824.

a:=n->sum(9^(n+1)*x^n/10^x,x=1..infinity): seq(a(n),n=0..17); # Emeric Deutsch, Mar 23 2007

a(n) = Sum_{x>=1} (9^(n+1))(x^n) / 10^x. - Alexander Walker, Feb 26 2007

Corrected and extended by Emeric Deutsch, Mar 23 2007

A099670 Partial sums of repdigits of A002277.

Original entry on oeis.org

3, 36, 369, 3702, 37035, 370368, 3703701, 37037034, 370370367, 3703703700, 37037037033, 370370370366, 3703703703699, 37037037037032, 370370370370365, 3703703703703698, 37037037037037031, 370370370370370364, 3703703703703703697, 37037037037037037030, 370370370370370370363

Offset: 1

Labos Elemer, Nov 17 2004

			3 + 33 + 333 + 3333 = a(4) = 3702.

Index entries for linear recurrences with constant coefficients, signature (12,-21,10).

Cf. A002275-A002283, A014824, A057932.

a:=n->sum((10^(n-j)-1^(n-j))/3,j=0..n): seq(a(n), n=1..18); # Zerinvary Lajos, Jan 15 2007

Mathematica
```
<Robert G. Wilson v, Nov 20 2004 *)
```

a(n) = (3/81)*(10^(n+1) - 9*n - 10). - R. Piyo (nagoya314(AT)yahoo.com), Dec 10 2004
From Elmo R. Oliveira, Apr 02 2025: (Start)
G.f.: 3*x/((1 - x)^2*(1 - 10*x)).
E.g.f.: 3*exp(x)*(10*exp(9*x) - 9*x - 10)/81.
a(n) = 3*A014824(n).
a(n) = 12*a(n-1) - 21*a(n-2) + 10*a(n-3) for n > 3. (End)

More terms from Elmo R. Oliveira, Apr 02 2025

A099671 Partial sums of repdigits of A002278.

Original entry on oeis.org

4, 48, 492, 4936, 49380, 493824, 4938268, 49382712, 493827156, 4938271600, 49382716044, 493827160488, 4938271604932, 49382716049376, 493827160493820, 4938271604938264, 49382716049382708, 493827160493827152, 4938271604938271596, 49382716049382716040, 493827160493827160484

Offset: 1

Labos Elemer, Nov 17 2004

			4 + 44 + 444 + 4444 + 44444 = a(5) = 49380.

Index entries for linear recurrences with constant coefficients, signature (12,-21,10).

Cf. A002275-A002283, A014824, A057932.

Mathematica
```
<Robert G. Wilson v, Nov 20 2004 *)
```

a(n) = (4/81)*(10^(n+1) - 9*n - 10). - R. Piyo (nagoya314(AT)yahoo.com), Dec 10 2004
From Chai Wah Wu, Feb 28 2018: (Start)
a(n) = 12*a(n-1) - 21*a(n-2) + 10*a(n-3) for n > 3.
G.f.: 4*x/((1 - x)^2*(1 - 10*x)). (End)
From Elmo R. Oliveira, Apr 03 2025: (Start)
E.g.f.: 4*exp(x)*(10*exp(9*x) - 9*x - 10)/81.
a(n) = 4*A014824(n). (End)

More terms from Elmo R. Oliveira, Apr 03 2025

cp's OEIS Frontend

A178643 Square array read by antidiagonals. Convolution of a(n) = 2*a(n-1) - a(n-2) and 10^n.

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A180027 Partial sums of A100706.

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A180039 Combinatorial tetrahedral numbers.

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A249181 a(n) = A057137(n)^2 where A057137 = 0,1,12,123,...,123...90,...

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A332082 a(n) = Sum_{1 <= m <= n} Sum_{1 <= k <= n+1-m} m*R(k,n+1), where R(k,b) = (b^k - 1)/(b - 1) is the base-b repunit of length k.

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A359846 a(n) = (((5 - (n mod 2))*10^(3 + n*(9/2) - (n mod 2)*(5/2)))^2 + 2)/81.

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A055528 a(n) = 10*a(n-1)+n^3, a(0)=0.

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A055530 The recurrence b(k) = 10*b(k-1) + k^n with b(0)=0 has b(k)/10^k converging to a(n)/9^(n+1).

A359846 a(n) = (((5 - (n mod 2))10^(3 + n(9/2) - (n mod 2)*(5/2)))^2 + 2)/81.