A037156
a(n) = 10^n*(10^n+1)/2.
Original entry on oeis.org
1, 55, 5050, 500500, 50005000, 5000050000, 500000500000, 50000005000000, 5000000050000000, 500000000500000000, 50000000005000000000, 5000000000050000000000, 500000000000500000000000, 50000000000005000000000000, 5000000000000050000000000000
Offset: 0
From _Omar E. Pol_, May 03 2015: (Start)
For n = 0; a(0) = 1 = 1 * 1 = 1
For n = 1; a(1) = 1 + 2 + ...... + 9 + 10 = 11 * 5 = 55
For n = 2; a(2) = 1 + 2 + .... + 99 + 100 = 101 * 50 = 5050
For n = 3; a(3) = 1 + 2 + .. + 999 + 1000 = 1001 * 500 = 500500
...
(End)
- C. Pickover, Wonders of Numbers, Oxford University Press, NY, 2001, p. 328.
A subsequence of the triangular numbers
A000217.
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LinearRecurrence[{110,-1000},{1,55},20] (* Harvey P. Dale, Oct 11 2023 *)
A238237
Numbers which when chopped into two parts with equal length, added and squared result in the same number.
Original entry on oeis.org
81, 2025, 3025, 9801, 494209, 998001, 24502500, 25502500, 52881984, 60481729, 99980001, 6049417284, 6832014336, 9048004641, 9999800001, 101558217124, 108878221089, 123448227904, 127194229449, 152344237969, 213018248521, 217930248900, 249500250000, 250500250000
Offset: 1
2025 = (20 + 25)^2, so 2025 is in the sequence.
3025 = (30 + 25)^2, so 3025 is in the sequence.
9801 = (98 + 01)^2, so 9801 is in the sequence.
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Select[Range[600000]^2, EvenQ[len=IntegerLength[#]] && # == (Mod[#,10^(len/2)] + Floor[#/10^(len/2)])^2 &] (* Stefano Spezia, Jan 01 2025 *)
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forstep(m=1, 7, 2, p=10^((m+1)/2); for(n=10^m, 10^(m+1)-1, d=lift(Mod(n, p)); if(((n-d)/p+d)^2==n, print1(n, ", "))));
A350870
Numbers k = x.y such that x.y = (x+y)^2, when x and y have the same number of digits, "." means concatenation, and y may not begin with 0.
Original entry on oeis.org
81, 2025, 3025, 494209, 24502500, 25502500, 52881984, 60481729, 6049417284, 6832014336, 101558217124, 108878221089, 123448227904, 127194229449, 152344237969, 213018248521, 217930248900, 249500250000, 250500250000, 284270248900, 289940248521, 371718237969, 413908229449, 420744227904
Offset: 1
81 = (8+1)^2, hence 81 is a term.
3025 = (30+25)^2, hence 3025 is another term.
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upto(n) = {i = 4; i2 = i^2; res = List(); while(i2 <= n, i++; i2 = i^2; if(#digits(i2) % 2 == 1, i = sqrtint(10^(#digits(i2))) + 1; i2 = i^2; ); if(is(i2), listput(res, i2) ); ); res }
is(n) = { my(d = digits(n), d1, d2, frd2); if(#d % 2 == 1, return(0)); d1 = vector(#d \ 2, i, d[i]); d2 = vector(#d \ 2, i, d[i + #d \ 2]); frd2 = fromdigits(d2); 10^(#d \ 2 - 1) <= frd2 && (fromdigits(d1) + frd2)^2 == n } \\ David A. Corneth, Jan 21 2022
A350869
a(n) = Sum_{i=0..10^n-1} i^3.
Original entry on oeis.org
0, 2025, 24502500, 249500250000, 2499500025000000, 24999500002500000000, 249999500000250000000000, 2499999500000025000000000000, 24999999500000002500000000000000, 249999999500000000250000000000000000, 2499999999500000000025000000000000000000
Offset: 0
a(1) = Sum_{i=0..9} i^3 = (Sum_{i=0..9} i)^2 = 2025.
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a[n_] := (10^n*(10^n - 1)/2)^2; Array[a, 11, 0] (* Amiram Eldar, Jan 20 2022 *)
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a(n) = my(x=10^n-1); (x*(x+1)/2)^2; \\ Michel Marcus, Jan 22 2022
Showing 1-4 of 4 results.
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