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A086639
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Write decimal expansion of Pi in triangular form; sequence gives left edge.
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4
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3, 1, 1, 2, 5, 3, 2, 2, 4, 9, 9, 7, 8, 3, 8, 7, 2, 1, 8, 9, 5, 3, 6, 6, 3, 5, 7, 6, 2, 2, 9, 9, 4, 0, 4, 2, 3, 0, 4, 1, 6, 7, 8, 9, 9, 1, 2, 3, 0, 1, 7, 2, 2, 4, 7, 8, 3, 1, 8, 3, 0, 2, 7, 9, 1, 6, 2, 2, 6, 7, 6, 8, 1, 5, 7, 3, 7, 7, 2, 4, 9, 3, 2, 1, 9, 8, 9, 1, 2, 7, 7, 9, 4, 0, 9, 2, 9, 8, 4, 9, 9, 2, 0, 7, 0
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OFFSET
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1,1
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COMMENTS
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In the second formula, "if" can most probably be strengthened to "if and only if": Indeed, a(n) = 0 can be equal to A000030(A090897(n)) only if A090897(n) = 0, i.e., there would be a string of n consecutive zeros in the decimals of Pi from position T(n-1)+1 to position T(n). The probability that this happens appears to be zero. (Notice how A096764(n), first occurrence of n consecutive zeros, grows incredibly much faster than T(n).) Maybe this could be proved considering, e.g., a continued fraction expansion of Pi whose coefficients follow some pattern of moderate growth (as e.g. in A046126), while a very long string of zeros in the decimal expansion would mean that it is exceptionally close to the rational number given by the truncation. - M. F. Hasler, Jan 06 2023
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LINKS
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FORMULA
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EXAMPLE
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Triangle is
3
14
159
2653
58979
323846
2643383
27950288
419716939
9375105820
a(34) = 0 because in the decimals of Pi there is a 0 at position 562, following the triangular number A000217(33) = 561, i.e., in the first column of the 34th row in the above triangle. - Michel Marcus and M. F. Hasler, Jan 06 2023
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MATHEMATICA
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pi = RealDigits[Pi, 10, 5461][[1]]; Table[ pi[[n(n + 1)/2 + 1]], {n, 0, 104}]
Module[{nn=110, pid}, pid=RealDigits[Pi, 10, (nn(nn+1))/2][[1]]; TakeList[ pid, Range[ nn]]][[;; , 1]] (* Harvey P. Dale, Mar 06 2023 *)
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CROSSREFS
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KEYWORD
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easy,nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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